Question 3015 Marks
Evaluate the following intregals:
$\int\frac{1}{1+3\sin^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{1+3\sin^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{\frac{1}{\cos^2\text{x}}+\frac{3\sin^2\text{x}}{\cos^2\text{x}}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{\sec^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+\tan^2\text{x}+3\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+4\tan^2\text{x}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{1+(2\tan\text{x})}\ \text{dx}$
Let $2\tan\text{x}=\text{t}$
$2\sec^2\times\text{dx}=\text{dt}$
$\text{I}=\frac{1}{2}\int\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\tan^{-1}\text{t}+\text{c}$
$\text{I}=\frac{1}{2}\tan^{-1}(2\tan\text{x})+\text{C}$
View full question & answer→Question 3025 Marks
Evaluate the following integrals:
$\int\tan\text{x}\sec^2\text{x}\sqrt{1-\tan^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\tan\text{x}\sec^2\text{x}\sqrt{1-\tan^2\text{x}}\text{ dx}\ ....(1)$ Let $1-\tan^2\text{x}=\text{t}$ then, $\Rightarrow\text{d}\big(1-\tan^2\text{x}\big)\text{dt}$ $\Rightarrow-2\tan\text{x}\sec^2\text{x}\text{ dx}=\text{dt}$ $\Rightarrow\tan\text{x}\sec^2\text{x}\text{ dx}=\frac{-\text{dt}}{2}$Putting $1-\tan^2\text{x}=\text{t}$ and $\tan\text{x}\sec^2\text{x}\text{ dx}=-\frac{\text{dt}}{2}$ in equation (1),
We get
$\text{I}=\int\sqrt{\text{t}}\times\frac{-\text{dt}}{2}$
$=\frac{-1}{2}\int\text{t}^{\frac{1}{2}}\text{dt}$
$=-\frac{1}{2}\times\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+\text{C}$
$=-\frac{1}{3}\text{t}^\frac{3}{2}+\text{C}$
$=-\frac{1}{3}\big[1-\tan^2\text{x}\big]^{\frac{3}{2}}+\text{C}$
View full question & answer→Question 3035 Marks
Evaluate the following integrals:
$\int\frac{\cos^3\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
Answer$\int\frac{\cos^3\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$=\int\frac{\cos^2\text{x}\cos\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$=\int\frac{(1-\sin^2\text{x})\cos\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$\text{Let }\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\text{Now,}\int\frac{(1-\sin^2\text{x})\cos\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$=\int\frac{(1-\text{t}^2)}{\sqrt{\text{t}}}\text{dt}$
$=\int\Big(\frac{1}{\sqrt{\text{t}}}-\text{t}^\frac{3}{2}\Big)\text{dt}$
$=\int\Big(\text{t}^{-\frac{1}{2}}-\text{t}^\frac{3}{2}\Big)\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}-\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}-\frac{2}{5}\text{t}^\frac{5}{2}+\text{C}$
$=2\sqrt{\sin\text{x}}-\frac{2}{5}\sin^\frac{5}{2}\text{x}+\text{C}$
View full question & answer→Question 3045 Marks
Integrate the function in Exercise:
$\frac{1}{\text{x}^{2}(\text{x}^{2}+1)^{\frac{3}{4}}}$
Answer$\frac{1}{\text{x}^{2}(\text{x}^{2}+1)^{\frac{3}{4}}}$Multipling and dividing by $\text{x}^{-3}$, we obtain
$\frac{\text{x}^{-3}}{\text{x}^{2}.\text{x}^{-3}(\text{x}^{4}+1)^{\frac{3}{4}}}=\frac{\text{x}^{-3}(\text{x}^{4}+1)^{\frac{3}{2}}}{\text{x}^{2}.\text{x}^{-3}}$
$=\frac{(\text{x}^{4}+1)^{\frac{-3}{4}}}{\text{x}^{5}.(\text{x)}^{4^{-\frac{3}{4}}}}$
$=\frac{1}{\text{x}^{5}}\bigg(\frac{\text{x}^{4}+1}{\text{x}^{4}}\bigg)^{-\frac{3}{4}}$
$\text{Let}\frac{1}{\text{x}^{4}}=\text{t} \Rightarrow-\frac{4}{\text{x}^{5}}\text{dx}=\text{dt}\Rightarrow\frac{1}{\text{x}^{5}}\text{dx}=-\frac{\text{dt}}{4}$
$\therefore\int\frac{1}{\text{x}^{2}(\text{x}^{4}+1)^{\frac{3}{4}}}\text{dx}=\int\frac{1}{\text{x}^{5}}\bigg(1+\frac{1}{\text{x}^{4}}\bigg)^{-\frac{3}{4}}\text{dx}$
$=-\frac{1}{4}\int(1+\text{t)}^{-\frac{3}{4}}\text{dt}$
$=-\frac{1}{4}\Bigg[\frac{(1+\text{t)}^{-\frac{3}{4}}}{\frac{1}{4}}\Bigg]+\text{C}$
$=-\frac{1}{4}\frac{\bigg(1+\frac{1}{\text{x}^{4}}\Bigg)^{\frac{1}{4}}}{\frac{1}{4}}+\text{C}$
$=-\Bigg(1+\frac{1}{\text{x}^{4}}\Bigg)^{\frac{1}{4}}+\text{C}$
View full question & answer→Question 3055 Marks
$\int\sin\text{x}\sqrt{1-\cos2\text{x}}\text{ dx}$
AnswerLet I $=\int\sin\text{x}\sqrt{1-\cos2\text{x}}\text{ dx}.$ Then,
$\text{I}=\int\sin\text{x}\times\sqrt{2\sin^2\text{x}}\times\text{dx}$
$=\int\sin\text{x}\times\sqrt{2}\times\sin\text{x dx}$
$=\sqrt{2}\int\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\int2\sin^2\text{x}\times\text{dx}$
$=\frac{\sqrt{2}}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]+\text{C}$
$=\frac{\sqrt{2}\text{x}}{2}-\frac{\sqrt{2}}{4}\times\sin2\text{x}+\text{C}$
$=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
$\therefore\text{I}=\frac{1}{\sqrt{2}}\times\text{x}-\frac{\sin2\text{x}}{2\sqrt{2}}+\text{C}$
View full question & answer→Question 3065 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{m}\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{\text{m}\sin^{-1}\text{x}}}{\sqrt{1-\text{x}^2}}\text{ dx}\ ....(1)$ Let $\text{m}\sin^{-1}\text{x}=\text{t}$ then, $\text{d}\big(\text{m}\sin^{-1}\text{x}\big)=\text{dt}$ $\Rightarrow\text{m}\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$ $\Rightarrow\frac{\text{dx}}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{m}}$Putting, $\text{m}\sin^{-1}\text{x}=\text{t}$ and $\frac{\text{dx}}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{m}}$ in equation (1),
We get,
$\text{I}=\int\text{e}^\text{t}\frac{\text{dt}}{\text{m}}$
$=\frac{1}{\text{m}}\text{e}^\text{t}+\text{C}$
$=\frac{1}{\text{m}}\text{e}^{\text{m}\sin^{-1}\text{x}}+\text{C}$
View full question & answer→Question 3075 Marks
If f(2a - x) = -f(x), prove that $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=0$
AnswerLet $\text{I}=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=0$
Using additive property
$\text{I}=\int\limits^{\text{a}}_0\text{f(x)}\text{dx}+\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Consider the integral $\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}$
Let $\text{x}=2\text{a}-\text{t},$ Then $\text{dx}=-\text{dt}$
When $\text{x}=\text{a},\text{t}=\text{a}$ and $\text{x}=2\text{a},\text{t}=0$
Threrfore,
$=\int\limits^{2\text{a}}_0\text{f(x)}\text{dx}=-\int\limits^0_\text{a}\text{f}(2\text{a}-\text{t})\text{dt}$
$=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{t})\text{dt}$
$=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$ (Chang ing the variable)
We have
$\text{f}(2\text{a}-\text{x})=-\text{f(x)}$
Therefore,
$\text{I}=\int\limits^\text{a}_0\text{f(x)}\text{dx}-\int\limits^\text{a}_0\text{f(x)}\text{dx}=0$
View full question & answer→Question 3085 Marks
Find the integrals of the functions in Exercises:
$\sin\text{x } \sin2\text{x }\sin3\text{x}$
AnswerIt is known thtat, $\sin\text{A}\sin\text{B}=\frac{1}{2}\big\{\cos(\text{A}-\text{B})-\cos(\text{A}+\text{B})\big\}$
$\therefore\int\sin\text{x }\sin2\text{x }\sin3\text{x}\text{ dx}$
$=\int\bigg[\sin\text{x}\cdot\frac{1}{2}\big\{\cos(2\text{x}-3\text{x})-\cos(2\text{x}+3\text{x})\big\}\bigg]\text{dx}$
$=\frac{1}{2}\int\big(\sin\text{x}\cos(-\text{x})-\sin\text{x}\cos5\text{x}\big)\text{dx}$
$=\frac{1}{2}\int\big(\sin\text{x}\cos\text{x}-\sin\text{x}\cos5\text{x}\big)\text{dx}$
$=\frac{1}{2}\int\frac{\sin2\text{x}}{2}\text{ dx}-\frac{1}{2}\int\sin\text{x}\cos5\text{x}\text{ dx}$
$=\frac{1}{4}\bigg[\frac{-\cos2\text{x}}{2}\bigg]-\frac{1}{2}\int\bigg\{\frac{1}{2}\sin(\text{x}+5\text{x})+\sin(\text{x}-5\text{x})\bigg\}\text{ dx}$
$=\frac{-\cos2\text{x}}{8}-\frac{1}{4}\int\big(\sin6\text{x}+\sin(-4\text{x})\big)\text{ dx}$
$=\frac{-\cos2\text{x}}{8}-\frac{1}{4}\bigg[\frac{-\cos6\text{x}}{3}+\frac{\cos4\text{x}}{4}\bigg]+\text{C}$
$=\frac{-\cos2\text{x}}{8}-\frac{1}{8}\bigg[\frac{-\cos6\text{x}}{3}+\frac{\cos4\text{x}}{2}\bigg]+\text{C}$
$=\frac{1}{8}\bigg[\frac{\cos6\text{x}}{3}-\frac{\cos4\text{x}}{2}-\cos2\text{x}\bigg]+\text{C}$
View full question & answer→Question 3095 Marks
Evaluate the following intregals: $\int\frac{1}{\sqrt{3}\sin\text{x}+\cos\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{3}\sin\text{x}+\cos\text{x}}\ \text{dx}$
Let $\sqrt{3}=\text{r}\cos\theta,\text{and }1=\text{r}\sin\theta$
$\tan\theta=\frac{1}{\sqrt{3}}$
$\theta=\frac{\pi}{6}$
$\text{r}=\sqrt{3+1}=2$
$\text{I}=\int\frac{1}{\text{r}\cos\theta\sin\text{x}+\text{r}\sin\theta\cos\text{x}}\ \text{dx}$
$=\frac{1}{\text{r}}\int\frac{1}{\sin(\text{x}+\theta)}\text{dx}$
$=\frac{1}{2}\int\text{cosec}(\text{x}+\theta)\text{dx}$
$=\frac{1}{2}\log\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\theta}{2}\Big)\Big|+\text{c}$
$\text{I}=\frac{1}{2}\log\Big|\tan\Big(\frac{\text{x}}{2}+\frac{\pi}{12}\Big)\Big|+\text{C}$
View full question & answer→Question 3105 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
AnswerWE have,
$\text{I}=\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{2\text{x}}\Big(\frac{1-2\sin\text{x}\cos\text{x}}{2\sin^2\text{x}}\Big)\text{dx}$
Put $\text{t}=2\text{x}.$ Then $\text{dt}=2\text{dx}$
Therefore,
$\text{I}=\frac{1}{2}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{2\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}4{\int\text{e}^{\text{t}}}\bigg(\frac{1}{\sin^2\frac{\text{t}}{2}}-\frac{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\big(\text{cosec}^2\frac{\text{t}}{2}-2\cot\frac{\text{t}}{2}\big)\text{dt}$
$=-\frac{1}{4}\int\text{e}^{\text{t}}\big(2\cot\frac{\text{t}}{2}-\text{cosec}^2\frac{\text{t}}{2}\big)\text{dt}$
Consider, $\text{f(x)}=2\cot\frac{\text{t}}{2},$ then $\text{f}'\text{(x)}=-\text{cosec}^2\frac{\text{t}}{2}$
Thus, the given integrand is of the from $\text{e}^{\text{x}}\big[\text{f(x)}+\text{f'}\text{(x)}\big].$
Therefore, $\text{I}=-\frac{1}{4}\big(2\cot\frac{\text{t}}{2}\big)\text{e}^{\text{t}}+\text{C}$
$=-\frac{1}{4}\big(2\cot\frac{2\text{x}}{2}\big)\text{e}^{2\text{x}}+\text{C}$
Hence, $\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}=-\frac{1}{2}(\cot\text{x})\text{e}^{2\text{x}}+\text{C}$
View full question & answer→Question 3115 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}$
AnswerWe have,
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}$
We have that,
$\int\frac{1}{\text{a}^2+\text{x}^2}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}$
$\therefore\ \frac{1}{\text{b}^2}\int_{0}^\limits{\infty}\frac{1}{\big(\frac{\text{a}}{\text{b}}\big)^2+\text{x}^2}\text{ dx}=\frac{1}{\text{b}^2}\Big[\frac{\text{b}}{\text{a}}\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\Big[\tan^{-1}\Big(\frac{\text{bx}}{\text{a}}\Big)\Big]^{\infty}_0$
$=\frac{1}{\text{ab}}\big[\tan^{-1}\infty-\tan^{-1}0\big]$
$=\frac{1}{\text{ab}}\Big[\frac{\pi}{2}-0\Big]$
$=\frac{\pi}{2\text{ab}}$
$\int_{0}^\limits{\infty}\frac{1}{\text{a}^2+\text{b}^2\text{x}^2} \text{ dx}=\frac{\pi}{2\text{ab}}$
View full question & answer→Question 3125 Marks
Evaluate the following integrals:
$\int\limits^{1}_0\big|\text{x}\sin\pi\text{x}\big|\text{dx}$
AnswerFor $0<\text{x}<1,\text{ x}>0$ and $\sin\pi\text{x}>0\Rightarrow\text{x}\sin\pi\text{x}>0$
$\therefore\ \int\limits^{1}_0|\text{x}\sin\pi\text{x}|\text{ dx}=\int\limits^{1}_0\text{x}\sin\pi\text{x}\text{ dx}$
Let $\text{I}=\int\text{x}\sin\pi\text{x}\text{ dx}$
$=\text{x}\int\sin\pi\text{x}\text{ dx}-\int\Big(\frac{\text{d}}{\text{dx}}\text{x}\int\sin\pi\text{x dx}\Big)\text{dx}$
$=\text{x}\Big(\frac{-\cos\pi\text{x}}{\pi}\Big)-\int\Big(\frac{-\text{cos}\pi\text{x}}{\pi}\Big)\text{dx}$
$=\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}$
Applying the limits, we get
$\int\limits^{1}_0|\text{x}\sin\pi\text{x}|\text{ dx}=\Big[\frac{-\text{x}\cos\pi\text{x}}{\pi}+\frac{\sin\pi\text{x}}{\pi^2}\Big]^1_0$
$=\Big(\frac{-\cos\pi}{\pi}+\frac{\sin\pi}{\pi^2}\big)-(0+0)$
$=\frac{1}{\pi}+0-0$
$=\frac{1}{\pi}$
View full question & answer→Question 3135 Marks
Integrate the function in exercise.
$\text{x}\ \tan^{-1}\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\tan^{-1}\text{x dx}$
Taking $\tan^{-1}\text{x}$ as first function and x as second function and integrating by parts, we obtain.
$\text{I}=\tan^{-1}\text{x}\int\text{x} \ \text{dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}\Big)\int\text{x} \ \text{dx}\Bigg\}\text{dx}$
$=\tan^{-1}\text{x}\Big(\frac{\text{x}^2}{2}\Big)-\int\frac{1}{1+\text{x}^2}.\frac{\text{x}^2}{2}\text{dx}$
$=\frac{\text{x}^2\tan^{-1}\text{x}}{2}-\frac{1}{2}\int\frac{\text{x}^2}{1+\text{x}^2}\text{dx}$
$=\frac{\text{x}^2\tan^{-1}\text{x}}{2}-\frac{1}{2}\int\Bigg(\frac{\text{x}^2+1}{1+\text{x}^2}-\frac{1}{1-\text{x}^2}\Bigg)\text{dx}$
$=\frac{\text{x}^2\tan^{-1}\text{x}}{2}-\frac{1}{2}\int\Bigg(1-\frac{1}{1+\text{x}^2}\Bigg)\text{dx}$
$=\frac{\text{x}^2\tan^{-1}\text{x}}{2}-\frac{1}{2}(\text{x}-\tan^{-1}\text{x})+\text{C}$
$=\frac{\text{x}^2}{2}\tan^{-1}\text{x}-\frac{\text{x}}{2}+\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
View full question & answer→Question 3145 Marks
Evaluate the following integrals:
$\int\text{x}^3\tan^{-1}\text{x dx}$
Answer$\int\text{x}^3\tan^{-1}\text{x dx}$
$\int\text{x}^3\tan^{-1}\text{x dx}=\tan^{-1}\text{x}\int\text{x}^3\text{dx}-\Big(\int\frac{\text{d}\tan^{-1}\text{x}}{\text{dx}}\big(\int\text{x}^3\text{dx}\big)\text{dx}\Big)$
$=\tan^{-1}\text{x}\frac{\text{x}^4}{4}-\Big(\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}\Big)$
$=\tan^{-1}\text{x}\frac{\text{x}^4}{4}-\Big(\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}\Big)$
$\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}=\frac{1}{4}\Big[\int\frac{1}{1+\text{x}^2}\text{dx}+(\text{x}^2-1)\text{dx}\Big]$
$\int\frac{1}{1+\text{x}^2}\Big(\frac{\text{x}^4}{4}\Big)\text{dx}=\frac{1}{4}\Big[\tan^{-1}\text{x}+\frac{\text{x}^3}{3}-\text{x}\Big]$
$\int\text{x}^3\tan^{-1}\text{x dx}=\frac{\text{x}^4}{4}\tan^{-1}\text{x}-\frac{1}{4}\Big[\tan^{-1}\text{x}+\frac{\text{x}^3}{3}-\text{x}\Big]+\text{C}$
View full question & answer→Question 3155 Marks
Evaluate the following integrals:
$\int\frac{1}{4\text{x}^2+12\text{x}+5}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{4\text{x}^2+12\text{x}+5}\text{dx}$
$=\frac{1}{4}\int\frac{1}{\text{x}^2+3\text{x}+\frac{5}{4}}\text{dx}$
$=\frac{1}4{}\int\frac{1}{\text{x}^2+2\times\text{x}\times\big(\frac{3}{2}\big)+\big(\frac{3}{2}\big)^2-\big(\frac{3}{2}\big)^2+\frac{5}{4}}\text{dx}$
$\text{I}=\frac{1}{4}\int\frac{1}{\Big(\text{x}+\frac{3}{2}\Big)^2-1}\text{dx}$
Let $\Big(\text{x}+\frac{3}{2}\Big)=\text{t}\ \dots(1)$
$\Rightarrow\text{dx = dt}$
So,
$\text{I}=\frac{1}{4}\int\frac{1}{\text{t}^2-(1)^2}\text{dt}$
$\text{I}=\frac{1}{4}\times\frac{1}{2\times(1)}\log\bigg|\frac{\text{t}-1}{\text{t}+1}\bigg|+\text{C}$ $\Big[\text{Since,} \int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\log\bigg|\frac{\text{x}-\text{a}}{\text{x+a}}\bigg|+\text{C}\Big]$
$\text{I}=\frac{1}{8}\log\Bigg|\frac{\text{x}+\frac{3}{2}-1}{\text{x}+\frac{3}{2}+1}\Bigg|+\text{C}$ [using (1)]
$\text{I}=\frac{1}{8}\log\bigg|\frac{2\text{x}+1}{2\text{x}+5}\bigg|+\text{C}$
View full question & answer→Question 3165 Marks
Integrate the function in Exercise:
$(\sin^{-1}\text{x})^2$
AnswerLet $\text{I}=\int(\sin^{-1}\text{x})^21.\text{dx}$
Taking $(\sin^{-1} x)^2$ as first function and 1 as second function and integrating by parts, we obtain.
$\text{I}=(\sin^{-1}\text{x})^2\int1\text{dx}-\int\Bigg\{\Big(\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})^2\int1.\text{dx}\Bigg\}\text{dx}$
$=\text{x}(\sin^{-1}\text{x})^2-\int\frac{2\sin^{-1}\text{x }}{\sqrt{1-\text{x}^2}}.\text{x} \ \text{dx}$
$=\text{x}(\sin^{-1}\text{x})^2.\int\sin^{-1}\text{x}.\Bigg(\frac{-2\text{x }}{\sqrt{1-\text{x}^2}}\Bigg)\text{dx}$
$=\text{x}(\sin^{-1}\text{x})^2+\Bigg[\sin^{-1}\text{x}\int\frac{-2\text{x }}{\sqrt{1-\text{x}^2}}\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\sin^{-1}\text{x}\Bigg)\int\frac{-2\text{x}}{\sqrt{1-\text{x}^2}}\text{dx}\Bigg\}\text{dx}\Bigg]$
$=\text{x}(\sin^{-1}\text{x})^2+\Bigg[\sin^{-1}\text{x}-2.\sqrt{1-\text{x}^2}-\int\frac{1}{\sqrt{1-\text{x}^2}}.2\sqrt{1-\text{x}^2}\text{dx}\Bigg]$
$=\text{x}(\sin^{-1}\text{x})^2+2\sqrt{1-\text{x}^2}\sin^{-1}\text{x}-\int2\text{dx}$
$=\text{x}(\sin^{-1}\text{x})^2+2\sqrt{1-\text{x}^2}\sin^{-1}\text{x}-2\text{x}+\text{C}$
View full question & answer→Question 3175 Marks
Evaluate the following integrals:$\int\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$
Let $\text{x}=\tan\text{t}$
$\text{dx}=\sec^2\text{t dt}$
$\text{I}=\int\cos^{-1}\Big(\frac{1-\tan^2\text{t}}{1+\tan^2\text{t}}\Big)\sec^2\text{t dt}$
$=\int\cos^{-1}(\cos2\text{t})\sec^2\text{t dt}$
$=\int2\text{t}\sec^2\text{x dx}$
$=2\Big[\text{t}\int\sec^2\text{t dt}-\int(1\int\sec^2\text{t dt})\text{dt}\Big]$
$=2[\text{t}\tan^2\text{t}-\int\tan\text{t dt}]$
$=2[\text{t}\tan^2\text{t}-\log\sec\text{t}]+\text{C}$
$=2\Big[\text{x}\tan^{-1}\text{x}-\log\sqrt{1+\text{x}^2}\Big]+\text{C}$
$\text{I}=2\text{x}\tan^{-1}\text{x}-\log|1+\text{x}^2|+\text{C}$
View full question & answer→Question 3185 Marks
Evaluate the following integrals:
$\int(\tan^{-1}\text{x}^2)\text{x dx}$
AnswerLet $\text{I}=\int(\tan^{-1}\text{x}^2)\text{x dx}$
Let $\text{x}^2=\text{t}$
$2\text{x dx = dt}$
$\text{I}=\frac{1}{2}\int\tan^{-1}\text{t dt}$
$=\frac{1}{2}\int1\tan^{-1}\text{t dt}$
$=\frac{1}{2}\Big[\tan^{-1}\text{t}\int\text{dt}-\Big(\int\frac{1}{1+\text{t}^2}\int\text{dt}\Big)\text{dt}\Big]$
$=\frac{1}{2}\Big[\text{t}\tan^{-1}\text{t}-\int\frac{\text{t}}{1+\text{t}^2}\text{dt}\Big]$
$=\frac{1}{2}\text{t}\tan^{-1}\text{t}-\frac{1}{4}\int\frac{2\text{t}}{1+\text{t}^2}\text{dt}$
$=\frac{1}{2}\text{t}\tan^{-1}\text{t}-\frac{1}{4}\log\big|1+\text{t}^2\big|+\text{C}$
$\text{I}=\frac{1}{2}\text{x}^2\tan^{-1}\text{x}^2-\frac{1}{4}\log\big|1+\text{x}^4\big|+\text{C}$
View full question & answer→Question 3195 Marks
Integrate the function in Exercise:
$\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}$
Answer$\text{Let I}=\int\frac{\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx} \ \ \ \ ...\text{(i)}$
$\text{Let Linear}=\text{A}\frac{\text{d}}{\text{dx}}(\text{Quadratic})+\text{B}$
$\Rightarrow\ \ \text{x}+2=\text{A}\frac{\text{d}}{\text{dx}}(\text{x}^2+2\text{x}+3)+\text{B}$
$\Rightarrow\ \ \text{x}+2=\text{A}(2\text{x}+2)+\text{B}\ \ \ \ ...\text{(ii)}$
$\Rightarrow\ \ \text{x}+2=2\text{Ax}+2\text{A}+\text{B}$
Comparing coefficients of x,
$2\text{A}=1\Rightarrow\ \ \text{A}=\frac{1}{2}$
Comparing constants,
$2\text{A}+\text{B}=2$
On solving, we get
$\text{A}=\frac{1}{2}, \ \ \text{B}=1$
Putting the values of A and B in eq. (ii),
$\text{x}+2=\frac{1}{2}(2\text{x}+2)+1$
Putting this value of x + 2 in eq. (i),
$\text{I}=\int\frac{\frac{1}{2}(2\text{x}+2)+1}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}$
$\text{I}=\frac{1}{2}\int\frac{2\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}+\int\frac{1}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}$
$\Rightarrow\ \ \text{I}=\frac{1}{2}\text{I}_1+\text{I}_2\ \ \ \ \ \ ...\text{(iii)}$
$\text{Now }\text{ I}_1=\int\frac{2\text{x}+2}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}$
$\text{Putting }\text{ x}^2+2\text{x}+3=\text{t}\ \ \Rightarrow\ \ \ 2\text{x}+2=\frac{\text{dt}}{\text{dx}}\ \ \Rightarrow\ \ \ (2\text{x}+2)\text{ dx}=\text{dt}$
$\therefore\ \ \text{I}_1=\int\frac{\text{dt}}{\sqrt{t}}=\int\text{t}^{\frac{-1}{2}}\text{ dt}=\frac{\text{t}^{\frac{1}{2}}}{\frac{1}{2}}$
$=2\sqrt{\text{t}}=2\sqrt{\text{x}^2+2\text{x}+3} \ \ \ \ ...\text{(iv)}$
$\text{Again I}_2=\int\frac{1}{\sqrt{\text{x}^2+2\text{x}+3}}\text{ dx}$
$=\int\frac{1}{\sqrt{\text{x}^{2}+2\text{x}+1+2}}=\int\frac{1}{\sqrt{(\text{x}+1)^2+(\sqrt{2})^2}}\text{ dx}$
$=\log\begin{vmatrix}\text{x}+1+\sqrt{(\text{x}+1)^2+(2)^2}\end{vmatrix}$
$=\log\begin{vmatrix}\text{x}+1+\sqrt{\text{x}^2+2\text{x}+3}\end{vmatrix} \ \ \ \ ...\text{(v)}$
Putting values of $I_1$ and $I_2$ in eq. (iii),
$\text{I}=\sqrt{\text{x}^2+2\text{x}+3}+\log\begin{vmatrix}\text{x}+1+\sqrt{\text{x}^2+2\text{x}+3}\end{vmatrix}+\text{c}$
View full question & answer→Question 3205 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin\text{x}}{1+\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\frac{\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\frac{1+\sin\text{x}-1}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{(1-\sin\text{x})}{(1+\sin\text{x})(1-\sin\text{x})}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{(1-\sin\text{x})}{1-\sin}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\big(\sec^2\text{x}-\sec\text{x}\tan\text{x}\big)\text{dx}$
$=\pi\big[\text{x}\big]^{\pi}_0-\pi\big[\tan\text{x}-\sec\text{x}\big]^{\pi}_0$
$=\pi^2-\pi(0+1-0+1)$
$=\pi^2-2\pi$
Hence, $\text{I}=\pi\Big(\frac{\pi}{2}-1\Big)$
View full question & answer→Question 3215 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\tan^{-1}\text{x dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$ Then,
$\text{I}=\int_{0}^\limits{1}\tan^{-1}\text{x dx}$
Integrating by parts,
$\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\int_{0}^\limits{1}\frac{\text{x}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\big[\text{x}\tan^{-1}\text{x}\big]^1_0-\frac{1}{2}\big[\log\big(\text{x}^2+1\big)\big]^1_0$
$\Rightarrow\text{I}=\frac{\pi}{4}-0-\frac{1}{2}\log2+0$
$\Rightarrow\text{I}=\frac{\pi}{4}-\frac{1}{2}\log2$
View full question & answer→Question 3225 Marks
Evaluvate the following intregals:
$\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}$
AnswerLet $\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}-2}+\frac{\text{C}}{\text{x}-4}$ $\Rightarrow1=\text{A}(\text{x}-2)(\text{x}-4)+\text{B}(\text{x})(\text{x}-4)+\text{Cx}(\text{x}-2)$Put x = 0
$\Rightarrow1=8\text{A}\Rightarrow\text{A}=\frac{1}{8}$ Put x = 2 $\Rightarrow1=-4\text{B}\Rightarrow\text{B}=-\frac{1}{4}$ Put x = 4 $\Rightarrow1=8\text{C}\Rightarrow\text{C}=\frac{1}{8}$ So, $\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}=\frac{1}{8}\int\frac{\text{dx}}{\text{x}}+\Big(-\frac{1}{4}\Big)\int\frac{\text{dx}}{\text{x}-2}+\frac{1}{8}\int\frac{\text{dx}}{\text{x}-4}$ $=\frac{1}{8}\log|\text{x}|-\frac{1}{4}\log|\text{x}-2|+\frac{1}{8}\log|\text{x}-4|+\text{C}$ $=\frac{1}{8}\log\Big|\frac{\text{x}(\text{x}-4)}{(\text{x}-2)^2}\Big|+\text{C}$ $\text{I}=\frac{1}{8}\log\Big|\frac{\text{x}(\text{x}-4)}{(\text{x}-2)^2}\Big|+\text{C}$
View full question & answer→Question 3235 Marks
$\int\frac{\text{x}-1}{\sqrt{\text{x}+4}}\ \text{dx}$
Answer$\text{Let I} =\int\Big(\frac{\text{x}-1}{\sqrt{\text{x}+4}}\Big)\text{dx}$
Putting x + 4 = t
Then, x = t - 4
Difference both sides
dx = dt
Now integral becomes,
$\text{I}=\int\Big(\frac{\text{t}-4-1}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\frac{\text{t}}{\sqrt{\text{t}}}-\frac{5}{\sqrt{\text{t}}}\Big)\text{dt}$
$=\int\Big(\text{t}^{\frac{1}{2}}-5\text{t}^{-\frac{1}{2}}\Big)\text{dt}$
$=\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}-5\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{3}\text{t}^\frac{3}{2}-10\sqrt{\text{t}}+\text{C}$
$=\frac{2}{3}(\text{x}+4)^\frac{3}{2}-10(\text{x}+4)^\frac{1}{2}+\text{C}$
View full question & answer→Question 3245 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\Big(\sqrt{\tan\text{x}}+\sqrt{\cot}\text{x}\Big)\text{dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\bigg(\sqrt{\frac{\sin\text{x}}{\cos\text{x}}}+\sqrt{\frac{\cos\text{x}}{\sin\text{x}}}\bigg)\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2\sin\text{x}\cos\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\sqrt{2}\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{1-(\sin\text{x}+\cos\text{x})^2}}\text{ dx}$
Let $\sin\text{x}-\cos\text{x}=\text{t}$ Then, $\cos\text{x}+\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1$ and $\text{x}=\frac{\pi}{4},\text{t}=0$
$\therefore\ \text{I}=\sqrt{2}\int^\limits0_{-1}\frac{\text{dt}}{\sqrt{1-\text{t}^2}}$
$\Rightarrow\text{I}=\sqrt{2}\Big[\sin^{-1}\text{t}\Big]^0_{-1}$
$\Rightarrow\text{I}=\frac{\pi}{\sqrt{2}}$
View full question & answer→Question 3255 Marks
Evaluate the following integrals:
$\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}$
Answer$\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1+\cos\text{x}}}{(1-\cos\text{x})^{\frac{3}{2}}}\text{ dx}\times\frac{\sqrt{1-\cos\text{x}}}{\sqrt{1-\cos\text{x}}}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sqrt{1-\cos^2\text{x}}}{(1-\cos\text{x})^2}\text{ dx}$
$=\int^\limits{\frac{\pi}{2}}_\frac{\pi}{3}\frac{\sin\text{x}}{(1-\cos\text{x})^2}\text{ dx}$
Let $1-\cos\text{x}=\text{t},$ Then $\sin\text{x dx}=\text{dt}$
When $\text{x}=\frac{\pi}{3},\text{ t}=\frac{1}{2}$ and $\text{x}=\frac{\pi}{2},\text{ t}=1$
Therefore the integral becomes
$=\int_{1}^{\frac{1}{2}}\frac{\text{dt}}{\text{t}^2}$
$=\Big[-\frac{1}{\text{t}}\Big]^1_\frac{1}{2}$
$=-1+2$
$=1$
View full question & answer→Question 3265 Marks
Evaluate the following integrals:
$\int\text{x}\sqrt{\text{x}^4+1}\text{dx}$
Answer$\text{I}=\int\text{x}\sqrt{\text{x}^4+1}\text{dx}$
$=\int\text{x}\sqrt{(\text{x}^2)^2+1}\text{dx}$
Putting $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx}=\text{dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
$\therefore\ \text{I}=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\int\sqrt{\text{t}^2+1}\text{dt}$
$=\frac{1}{2}\Big[\frac{\text{t}}{2}\sqrt{\text{t}^2+1}+\frac{1^2}{2}\log\big|\text{t}+\sqrt{\text{t}^2+1}\big|\Big]+\text{C}$
$=\frac{1}{2}\Big[\frac{\text{x}^2}{2}\sqrt{\text{x}^4+1}+\frac{1}{2}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}\big|\Big]+\text{C}$
$=\frac{\text{x}^2}{4}\sqrt{\text{x}^4+1}+\frac{1}{4}\log\big|\text{x}^2+\sqrt{\text{x}^4+1}+\text{C}$
View full question & answer→Question 3275 Marks
Evaluate the following integrals:
$\int^\limits1_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}$
AnswerLet $\text{x}=\cos2\theta$
Differentiating w.r.t. x, we get
$\text{dx}=-2\sin2\theta\text{ d}\theta$
Now, $\text{x}=0\Rightarrow\theta=\frac{\pi}{4}$
$\text{x}=1\Rightarrow\theta=0$
$\therefore\ \int^\limits1_0\sqrt{\frac{1-\text{x}}{1+\text{x}}}\text{ dx}=\int^0\limits_\frac{\pi}{4}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\big(-2\sin2\theta\big)\text{d}\theta$
$=\int_0\limits^\frac{\pi}{4}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\big(2\sin2\theta\big)\text{d}\theta$ $\Big[\because\sin2\theta=2\sin\theta\cos\theta;\text{ and }\sin^2\theta=\frac{1-\cos2\theta}{2}\Big]$
$=2=\int_0\limits^\frac{\pi}{4}\frac{\sin\theta}{\cos\theta}\cdot\sin2\theta\text{ d}\theta$
$=4\int_0\limits^\frac{\pi}{4}\sin^2\theta\text{ d}\theta$
$=2\int_0\limits^\frac{\pi}{4}\big(1-\cos2\theta\big)\text{d}\theta$
$=2\Big[\theta-\frac{\sin^2\theta}{2}\Big]^{\frac{\pi}{4}}_0$
$=2\Big[\frac{\pi}{4}-\frac{1}{2}\Big]$
$=\frac{\pi}{2}-1$
View full question & answer→Question 3285 Marks
Evaluate the following integrals:
$\int\frac{\cot\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
AnswerLet I $=\int\frac{\cot\text{x}}{\sqrt{\sin\text{x}}}\text{dx}\ .....(1)$
Let $\sin\text{x}=\text{t}$ then,
$\text{d}(\sin\text{x})=\text{dt}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
$\text{Now,}\text{I}=\int\frac{\cot\text{x}}{\sqrt{\sin\text{x}}}\text{dx}$
$=\int\frac{\cot\text{x}}{\sin\text{x}\sqrt{\sin\text{x}}}\text{dx}$
$\int\frac{\cos\text{x}}{(\sin\text{x})^{\frac{3}{2}}}\text{dx}$
$\Rightarrow\ =\int\frac{\cos\text{x}}{(\sin\text{x})^\frac{3}{2}}\text{dx}\ ...(2)$
Putting $\sin\text{x}=\text{t}$ and $\cos\text{x}\text{ dx}=\text{dt}$ in equation (2), we get
$\text{I}=\int\frac{\text{dt}}{\text{t}^\frac{3}{2}}$
$=\int\text{t}^{-\frac{3}{2}}\text{dt}$
$=-2\text{t}^{-\frac{1}{2}}+\text{C}$
$=\frac{-2}{\sqrt{\text{t}}}+\text{C}$
$=\frac{-2}{\sqrt{\sin\text{x}}}+\text{C}$
$\therefore\text{I}=\frac{-2}{\sqrt{\sin\text{x}}}+\text{C}$
View full question & answer→Question 3295 Marks
Evaluate the following integrals:$\int\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{x}=\tan\theta\Rightarrow\text{dx}=\sec^2\theta\text{d}\theta$
$\therefore\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)=\sin^{-1}\Big(\frac{2\tan\theta}{1+\tan^2\theta}\Big)=\sin^{-1}(\sin2\theta)=\theta$
$\Rightarrow\int\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)\text{dx}=\int2\theta\cdot\sec^2\theta\text{d}\theta=2\int\theta\cdot\sec^2\theta\text{d}\theta$
Integrating by parts, we obtain
$2\Big[\theta\cdot\int\sec^2\theta\text{d}\theta-\int\Big\{\Big(\frac{\text{d}}{\text{d}\theta}\theta\Big)\int\sec^2\theta\text{d}\theta\Big\}\text{d}\theta\Big]$
$=2\big[\theta\cdot\tan\theta-\int\tan\theta\text{d}\theta\big]$
$=2\big[\theta\tan\theta+\log|\cos\theta|\big]+\text{C}$
$=2\Big[\text{x}\tan^{-1}\text{x}+\log\Big|\frac{1}{\sqrt{1+\text{x}^2}}\Big|\Big]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\log(1+\text{x}^2)^{-\frac{1}{2}}+\text{C}$
$=2\text{x}\tan^{-1}\text{x}+2\Big[-\frac{1}{2}\log(1+\text{x}^2)\Big]+\text{C}$
$=2\text{x}\tan^{-1}\text{x}-\log(1+\text{x}^2)+\text{C}$
View full question & answer→Question 3305 Marks
Evaluate the following intregals:
$\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{4\sin^2\text{x}+5\cos^2\text{x}}\text{ dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{4\tan^2\text{x}+5}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{4\tan^2\text{x}+5}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\ \text{dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{4+9(\text{t})^2}$
$=\int\frac{\text{dt}}{4\text{t}^2+5}$
Let $2\text{t}=\text{u}$
$2\text{dt}=\text{du}$
$\text{I}=\frac{1}{2}\int\frac{\text{du}}{(4)^2+(\sqrt{5})^2}$
$=\frac{1}{2}\times\frac{1}{\sqrt{5}}\times\tan^{-1}\Big(\frac{\text{u}}{\sqrt{5}}\Big)+\text{C}$
$=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\text{t}}{\sqrt{5}}\Big)+\text{C}$
$\text{I}=\frac{1}{2\sqrt{5}}\tan^{-1}\Big(\frac{2\tan\text{x}}{\sqrt{5}}\Big)+\text{C}$
View full question & answer→Question 3315 Marks
Evaluate the following integrals:
$\int\limits^1_{-1}\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\limits^1_{-1}\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)\text{dx}$
Here $\text{f(x)}=\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)$
$\text{f}(-\text{x})=\log\Big(\frac{2+\text{x}}{2-\text{x}}\Big)$
$=-\log\Big(\frac{2-\text{x}}{2+\text{x}}\Big)$
$=-\text{f(x)}$
Hence f(x) is an odd function,
Therefore,
$\text{I}=0$
View full question & answer→Question 3325 Marks
Evaluate the following integrals:
$\int\tan^{-1}(\sqrt{\text{x}})\text{dx}$
AnswerLet $\text{I}=\int\tan^{-1}(\sqrt{\text{x}})\text{dx}$Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\text{I}=\int2\text{t}\tan^{-1}\text{t dt}$
$=2\Big[\tan^{-1}\text{t}\int\text{t dt}-\int\Big(\frac{1}{1+\text{t}^2}\int\text{t dt}\Big)\text{dt}\Big]$
$=2\Big[\frac{\text{t}^2}{2}\tan^{-1}\text{t}-\int\frac{\text{t}^2}{2(1+\text{t}^2)}\text{dt}\Big]$
$=\text{t}^2\tan^{-1}\text{t}-\int\frac{\text{t}^2+1-1}{1+\text{t}^2}\text{dt}$
$=\text{t}^2\tan^{-1}\text{t}-\int\Big(1-\frac{1}{1+\text{t}^2}\Big)\text{dt}$
$=\text{t}^2\tan^{-1}\text{t}-\text{t}+\tan^{-1}\text{t + C}$
$=(\text{t}^2+1)\tan^{-1}\text{t}-\text{t + C}$
$\text{I}=(\text{x}+1)\tan^{-1}\sqrt{\text{x}}-\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 3335 Marks
$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log(\sin\text{x}+\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log(\sin\text{x}+\cos\text{x})\text{dx}\ \ \dots(\text{i})$
$ \text{I}=\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\log\bigg\{\sin\Big(\frac{\pi}{4}-\frac{\pi}{4}-\text{x}\Big)+\cos\Big(\frac{\pi}{4}-\frac{\pi}{4}-\text{x}\Big)\bigg\}\text{dx}$
$ =\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log\{\sin(-\text{x})+\cos(-\text{x})\}\text{dx} $
and $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log(\cos\text{x}-\sin\text{x})\text{dx}\ \ \dots(\text{ii})$
From Eqs. (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log\cos2\text{x dx}$
$2\text{I}=\int\limits^{\frac{\pi}{4}}_{0}\log\cos2\text{x dx}\ \ \dots(\text{iii})$
$ \bigg[\because\ \int\limits^{\text{a}}_{-\text{a}}\text{f(x)dx}=2\int\limits^{\text{a}}_{0}\text{f(x), if f}(-\text{x})=\text{f(x)}\bigg]$
Put $2\text{x}=\text{t}\Rightarrow\ \text{dx}=\frac{\text{dt}}{2} $
As $\text{x}\rightarrow0,$ then $\text{t}\rightarrow0$
and $\text{x}\rightarrow\frac{\pi}{4}$ then $\text{t}\rightarrow\frac{\pi}{2}$
$2\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\cos\text{t dt}\ \ \dots(\text{iv})$
$2\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\cos\bigg(\frac{\pi}{2}-\text{t}\bigg)\text{ dt}$ $ \bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)dx}=\int\limits^{\text{a}}_{0}\text{f}(\text{a}-\text{x})\text{dx}\bigg] $
$ \Rightarrow\ 2\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{t dt}\ \ \dots(\text{v}) $
On adding Eqs. (iv) and (v), we get
$ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{t}\cos\text{t dt} $
$\Rightarrow\ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\frac{\sin2\text{t}}{2}\text{dt} $
$ \Rightarrow\ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\sin2\text{x dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log2\text{dx}$
$ \Rightarrow\ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\sin\Big(\frac{\pi}{2}-2\text{x}\Big)\text{ dx}-\log2\cdot\frac{\pi}{4} $
$ \Rightarrow\ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\cos2\text{x}\text{ dx}-\frac{\pi}{4}\log2$
$\Rightarrow\ 4\text{I}=\int\limits^{\frac{\pi}{4}}_{0}\log\cos2\text{x}\text{ dx}-\frac{\pi}{4}\log2$ $\bigg[\because\ \int\limits^{2\text{a}}_{0}\text{f(x)dx}=2\int\limits^{\text{a}}_{0}\text{f(x)dx}\bigg]$
$\Rightarrow\ 4\text{I}=2\text{I}-\frac{\pi}{4}\log2$ [from Eq.(iii)]
$\text{I}=\frac{\pi}{8}\log2\frac{\pi}{8}\log\bigg(\frac{1}{2}\bigg)$
View full question & answer→Question 3345 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}^2-1)\sqrt{\text{x}^2+1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{(\text{x}^2-1)\sqrt{\text{x}^2+1}}\text{ dx}$
Let $\text{x}=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=-\int\frac{\frac{1}{\text{t}^2}\text{ dt}}{\Big(\frac{1}{\text{t}^2}-1\Big)\sqrt{\Big(\frac{1}{\text{t}^2}+1\Big)}}$
$=-\int\frac{\text{t dt}}{(1-\text{t}^2)\sqrt{1+\text{t}^2}}$
Let $1+\text{t}^2=\text{u}^2$
$2\text{tdt}=2\text{udt}$
$\text{I}=\int\frac{\text{udu}}{(\text{u}^2-2)\text{u}}$
$=\int\frac{\text{du}}{\text{u}^2-2}$
$\therefore\ \text{I}=\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{u}-\sqrt{2}}{\text{u}+\sqrt{2}}\bigg|+\text{C}$
$=\frac{1}{2\sqrt{2}}\log\bigg|\frac{\sqrt{1+\text{t}^2}-\sqrt{2}}{\sqrt{1+\text{t}^2}+\sqrt{2}}\bigg|+\text{C}$
Hence,
$\text{I}=-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\sqrt{2}\text{x}+\sqrt{\text{x}^2+1}}{\sqrt{2}\text{x}-\sqrt{\text{x}^2+1}}\bigg|+\text{C}$
View full question & answer→Question 3355 Marks
Evaluate the following integral:
$\int\frac{1}{\sqrt{(2-\text{x})^2+1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{(2-\text{x})^2+1}}\text{ dx}$
Let $2-\text{x}=\text{t}$
$-\text{dx}=\text{dt}$
$\text{dx}=-\text{dt}$
So, $\text{I}=-\int\frac{1}{\sqrt{\text{t}^2+(1)^2}}\text{ dt}$
$\text{I}=-\log\big|\text{t}+\sqrt{\text{t}^2+1}\big|+\text{C}$ $\Big[$Since $\int\frac{1}{\sqrt{\text{x}^2+\text{a}^2}}\text{ dx}=\log\big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\big|+\text{C}\Big]$
$\text{I}=-\log\big|(2-\text{x})+\sqrt{(2-\text{x})^2+1}\big|+\text{C}$ $[$Since $\text{t}=(2-\text{x})]$
View full question & answer→Question 3365 Marks
Evaluate the following integrals:
$\int\limits^\pi_0\frac{\text{x}}{1+\sin\alpha\sin\text{x}}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^\pi_0\frac{\text{x}}{1+\sin\alpha\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^\pi_0\frac{\pi-\text{x}}{1+\cos\alpha\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^\pi_0\frac{\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^\pi_0\frac{\text{x}+\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{\pi}{2}\int\limits^\pi_0\frac{1}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{1}{1+\cos\alpha\sin\text{x}}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{1}{1+\cos\alpha\frac{2\tan\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}+2\cos\alpha\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^\pi_0\frac{\sec^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}+2\cos\alpha\tan^{2}\frac{\text{x}}{2}}\text{ dx}$
Putting $\tan\frac{\text{x}}{2}=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{ t}\rightarrow0$
and $\text{x}\rightarrow\pi;\text{ t}\rightarrow\infty$
$\therefore\ \text{I}=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{1+\text{t}^2+2\cos\alpha+1}\text{ dt}$
$=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{(\text{t}+\cos\alpha)-\cos^2\alpha+1}\text{ dt}$
$={\pi}\int\limits^{\infty}_0\frac{1}{(\text{t}+\cos\alpha)+\sin^2\alpha}\text{ dt}$
$=\pi\Big[\frac{1}{\sin\alpha}\tan^{-1}\Big(\frac{1+\cos\alpha}{\sin\alpha}\Big)\Big]^1_0$
$=\frac{\pi}{\sin\alpha}\Big[\tan^{-1}(\infty)-\tan^{-1}(\cot\alpha)\Big]$
$=\frac{\pi}{\sin\alpha}\Big[\frac{\pi}{2}-\tan^{-1}\Big(\tan\Big(\frac{\pi}{2}-\alpha\Big)\Big)\Big]$
$=\frac{\pi\alpha}{\sin\alpha}$
View full question & answer→Question 3375 Marks
Evaluate the following intergrals:
$\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}$
$\Rightarrow-\text{e}^\text{ax}\frac{\cos(\text{bx}+\text{x})}{\text{b}}+\int\text{ae}^\text{ax}\frac{\cos(\text{bx}+\text{c})}{\text{b}}\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\int\text{e}^\text{ax}\cos(\text{bx}+\text{c})\text{dx}$
$=-\frac{1}{\text{b}}\text{e}^\text{ax}\cos(\text{bx}+\text{c})+\frac{\text{a}}{\text{b}}\Big[\int\text{e}^\text{ax}\frac{\sin(\text{bx}+\text{c})}{\text{b}}-\int\text{ae}^{\text{ax}}\frac{\sin(\text{bx}+\text{c})}{\text{b}}\text{dx}\Big]+\text{C}_1$
$=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\int\text{e}^\text{ax}\sin(\text{bx}+\text{c})\text{dx}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}\\-\frac{\text{a}^2}{\text{b}^2}\text{I}+\text{C}_1$
$\Rightarrow\text{I}=\Big\{\frac{\text{a}^2+\text{b}^2}{\text{b}^2}\Big\}-\frac{\text{e}^\text{ax}}{\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{a}^2+\text{b}^2}\big\{\text{a}\sin(\text{bx}+\text{c})-\text{b}\cos(\text{bx}+\text{c})\big\}+\text{C}_1$
View full question & answer→Question 3385 Marks
Evaluate the follwing intregals:
$\int\frac{1}{\text{x}^4-1}\text{ dx}$
AnswerThe Evaluate the integral follow the steps,
$\int\frac{1}{(\text{x}^4-1)}\text{ dx}$
Let $\frac{1}{(\text{x}^4-1)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}^2+1}$
$1={\text{A}}{(\text{x}-1})(\text{x}^2+1)+\text{B}(\text{x}+1)(\text{x}^2+1)+\text{C}(\text{x}+1)(\text{x}-1)$
$\text{For x}=1,\text{ B}=\frac{1}{4}$
$\text{For x}=-1,\text{ A}=-\frac{1}{4}$
$\text{For x}=0,\text{ C}=-\frac{1}{2}$
Therefore,
$\int\frac{1}{(\text{x}^4-1)}\text{ dx}=-\frac{1}{4}\int\frac{\text{dx}}{\text{x}+1}+\frac{1}{4}\frac{\text{dx}}{\text{x}-1}-\frac{1}{2}\int\frac{\text{dx}}{\text{x}^2+1}$
$=-\frac{1}{4}\ln\big|(\text{x}+1)\big|+\frac{1}{4}\ln\big|(\text{x}-1)\big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
$=\frac{1}{4}\ln\Big|\frac{\text{x}-1}{\text{x}+1}\Big|-\frac{1}{2}\tan^{-1}\text{x}+\text{C}$
View full question & answer→Question 3395 Marks
Evaluate the following integrals as limit of sum:
$\int\limits^2_0(\text{x}+3)\text{dx}$
Answer$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}+3,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_0(\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}(0+(\text{n}-1)\text{h})\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+3)+(0+\text{h}+3)+\ ....\ +(0+(\text{n}-1)\text{h}+3)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[3\text{n}+\text{h}\{1+2+3+\ ....\ +(\text{n}-1)\}\Big]$
$=\lim\limits_{\text{n}\rightarrow0}\text{h}\Big[3\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[3\text{n}+\text{n}-1\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(4-\frac{1}{\text{n}}\Big)$
$=8$
View full question & answer→Question 3405 Marks
Evaluate the following integrals:
$\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\limits^{\pi}_0\text{x}\sin\text{x}\cos^4\text{x}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin(\pi-\text{x})\cos^4(\pi-\text{x})\text{dx}$
$=\int\limits^{\pi}_0(\pi-\text{x})\sin\text{x}\cos^4\text{x dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\sin\text{x}\cos^4\text{x dx}$
$=\pi\int\limits^{\pi}_0\sin\text{x}\cos^4\text{x dx}$
Let $\cos\text{x}=\text{t},$ Then $-\sin\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=1,\text{x}=\pi,\text{t}=-1$
Therefore, $2\text{I}=-\pi\int\limits^{-1}_1\text{t}^4\text{ dt}$
$=\pi\int\limits^{1}_{-1}\text{t}^4\text{ dt}$
$=\pi\Big[\frac{\text{t}^5}{5}\Big]^{1}_{-1}$
$=\frac{\pi}{5}+\frac{\pi}{5}$
$=\frac{2\pi}{5}$
Hence, $\text{I}=\frac{\pi}{5}$
View full question & answer→Question 3415 Marks
Evaluate the integral in Exercise:
$\int^{2}_{0}\frac{\text{dx}}{\text{x}+4-\text{x}^{2}}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{2}\frac{\text{dx}}{\text{x}+4-\text{x}^{2}}=\int\limits_{0}^{2}\frac{\text{dx}}{4-(\text{x}^{2}-\text{x})}$
$=\int\limits_{0}^{2}\frac{\text{dx}}{\text{x}+4-\text{x}^{2}}=\int\limits_{0}^{2}\frac{\text{dx}}{4-(\text{x}^{2}-\text{x})}$
$=\int\limits_{0}^{2}\frac{\text{dx}}{\bigg(4+\frac{1}{4}\bigg)-\bigg(\text{x}^{2}-\text{x}+\frac{1}{4}\bigg)}=\int\limits_{0}^{2}\frac{\text{dx}}{\bigg(\frac{\sqrt{17}}{2}\bigg)^{2}-\bigg(\text{x}-\frac{1}{2}\bigg)^{2}}$
$=\frac{1}{2\times\frac{\sqrt{17}}{2}}\begin{bmatrix} \log\left|\frac{\frac{\sqrt{17}}{2}+\bigg(\text{x}-\frac{1}{2}\bigg)}{\frac{\sqrt{17}}{2}-\bigg(\text{x}-\frac{1}{2}\bigg)}\right| \\ \end{bmatrix}^{2}_{0}$
$=\frac{1}{\sqrt17}\begin{bmatrix} \log\left\{\frac{\frac{\sqrt{17}}{2}+\bigg({2}-\frac{1}{2}\bigg)}{\frac{\sqrt{17}}{2}-\bigg({2}-\frac{1}{2}\bigg)}\right\}-\log\left\{\frac{\frac{\sqrt{17}}{2}-\frac{1}{2}}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right\} \\ \end{bmatrix}$
$={\frac{\sqrt{1}}{17}}\bigg[\log\frac{\sqrt{17+3}}{\sqrt{17}-3}-\log\frac{\sqrt{17}-1}{\sqrt{17}+1}\bigg]$
$={\frac{\sqrt{1}}{17}}\log\bigg[\frac{\sqrt{17+3}}{\sqrt{17}-3}\times\frac{\sqrt{17}+1}{\sqrt{17}-1}\bigg]=\frac{1}{\sqrt{17}}\log\bigg(\frac{17+3+4\sqrt{17}}{17+3+-\sqrt{17}}\bigg)$
$={\frac{\sqrt{1}}{17}}\log\bigg(\frac{5+\sqrt{17}}{5-\sqrt{17}}\bigg)=\frac{1}{17}\log\bigg(\frac{5+\sqrt{17}}{5-\sqrt{17}}\bigg)=\frac{1}{\sqrt{17}}\log\bigg(\frac{5+\sqrt{17}}{5-\sqrt{17}}\times\frac{5+\sqrt{17}}{5+\sqrt{17}}\bigg)$
$={\frac{\sqrt{1}}{17}}\log\bigg[\frac{25+17+10\sqrt{17}}{25-17}\bigg]=\frac{1}{\sqrt{17}}\log\bigg[\frac{42+10\sqrt{17}}{8}\bigg]$
$={\frac{\sqrt{1}}{17}}\log\bigg[\frac{21+5\sqrt{7}}{4}\bigg]$
View full question & answer→Question 3425 Marks
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{dx}}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ a},\text{b}>0$
Answer$\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\cos\text{x}+\text{b}\sin\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{1}{\text{a}\Bigg(\frac{1-\tan^{2}\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)+\text{b}\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^{2}\frac{\text{x}}{2}}\Bigg)}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\big(1+\tan^{2}\frac{\text{x}}{2}\big)}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sec^2\frac{\text{x}}{2}}{\text{a}-\text{a}\tan^{2}\frac{\text{x}}{2}+2\text{b}\tan\frac{\text{x}}{2}}\text{ dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$ Then, $\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{ dx}=\text{dt}$
When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore the integral becomes
$\text{I}=\int^\limits1_0\frac{2\text{dt}}{\text{a}-\text{a}\text{t}^2+2\text{bt}}$
$=\int^\limits1_0\frac{2\text{dt}}{-\text{a}\big[\text{t}^2-\frac{2\text{bt}}{\text{a}-1}\big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{-\Big[\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2-1-\frac{\text{b}^2}{\text{a}^2}\Big]}$
$=\frac{2}{\text{a}}\int^\limits1_0\frac{\text{dt}}{\Big(\frac{\text{b}^2}{\text{a}^2}+1\Big)-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)^2}$
$=\frac{2}{\text{a}}\begin{bmatrix}\frac{1}{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}}\begin{pmatrix}\log\begin{vmatrix}\frac{2\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}}+\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}{\sqrt{\frac{\text{a}^2+\text{b}^2}{\text{a}^2}-\big(\text{t}-\frac{\text{b}}{\text{a}}\big)}}\end{vmatrix} \end{pmatrix}^1_0\end{bmatrix}$
$=\frac{1}{\sqrt{\text{a}^2+\text{b}^2}}\log\bigg(\frac{\text{a}+\text{b}+\sqrt{\text{a}^2+\text{b}^2}}{\text{a}+\text{b}-\sqrt{\text{a}^2+\text{b}^2}}\bigg)$
View full question & answer→Question 3435 Marks
Evaluate the following integrals:$\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$
Answer$\text{I}=\int\frac{\text{x}^3}{\text{x}^4+\text{x}^2+1}\text{ dx}$ $=\int\frac{\text{x}^2\cdot\text{x}}{(\text{x}^2)^2+\text{x}^2+1}\text{ dx}$ Let $\text{x}^2=\text{t}$ or $2\text{x}\text{ dx}=\text{dt}$ $\text{I}=\frac{1}{2}\int\frac{\text{t}}{\text{t}^2+\text{t}+1}\text{ dt}$ $=\frac{1}{4}\int\frac{2\text{t}}{\text{t}^2+\text{t}+1}\text{ dt}$$=\frac{1}{4}\int\frac{2\text{t}+1-1}{\text{t}^2+\text{t}+1}\text{ dt}$
$=\frac{1}{4}\int\Big[\frac{(2\text{t}+1)}{(\text{t}^2+\text{t}+1)}-\frac{1}{(\text{t}^2+\text{t}+1)}\Big]\text{dt}$
$=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}\big)}\text{ dt}\Big]$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\int\frac{1}{\big(\text{t}+\frac{1}{2}\big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\text{ dt}\end{bmatrix}$ $=\frac{1}{4}\begin{bmatrix}\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\frac{\big(\text{t}+\frac{1}{2}\big)}{\Big(\frac{\sqrt3}{2}\Big)}\end{bmatrix}+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{t}^2+\text{t}+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{t}+1}{\sqrt3}\Big)\Big]+\text{C}$ $=\frac{1}{4}\Big[\log\big|\text{x}^4+\text{x}^2+1\big|-\frac{2}{\sqrt3}\tan\Big(\frac{2\text{x}^2+1}{\sqrt3}\Big)\Big]+\text{C}$
View full question & answer→Question 3445 Marks
Integrate the function in Exercise:
$\frac{1}{\cos(\text{x}+\text{a)}\cos(\text{x}+\text{b)}}$
Answer$\frac{1}{\cos(\text{x}+\text{a)}\cos(\text{x}+\text{b)}}$Multiplying and dividing by $\sin(\text{a}-\text{b)},$ we obtain
$\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin\text{(a}-\text{b})}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin[\text{(x}+\text{a})-\text{(x}+\text{b)]}}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin\text{(x}+\text{a}).\cos\text{(x}+\text{b)}-\cos\text{(x}+\text{a)}\sin\text{(x}+\text{b)}}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin\text{(x}+\text{a)}}{\cos\text{(x}+\text{a)}}-\frac{\sin\text{(x}+\text{b)}}{\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\big[\tan(\text{x}+\text{a)}-\tan\text{(x}+\text{b)}\big]$
$\int\frac{1}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\text{dx}=\frac{1}{\sin\text{(a}-\text{b)}}\int\big[\tan\text{(x}+\text{a)}-\tan\text{(x}+\text{b)}\big]\text{dx}$
$=\frac{1}{\sin\text{(a}-\text{b)}}[-\log|\cos\text{(x}+\text{a)}|+\log|\cos\text{(x}+\text{b)|]}+\text{C}$
$=\frac{1}{\sin\text{(a}-\text{b)}}\log\bigg|\frac{\cos\text{(x}+\text{b)}}{\cos\text{(x}+\text{a)}}\bigg|+\text{C}$
View full question & answer→Question 3455 Marks
Evaluate the following:
$\int\frac{\sin^{-1}\text{x}}{(1-\text{x}^2)^{\frac{3}{4}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin^{-1}\text{x}}{(1-\text{x}^2)^{\frac{3}{4}}}\text{dx}$ $=\int\frac{\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}\text{dx}$
Put $\sin^{-1}\text{x}=\text{t}\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
And $\text{x}-\sin\text{t}\Rightarrow1-\text{x}^2=\cos^2\text{t}$
$\cos\text{t}=\sqrt{1-\text{x}^2}$
$\text{I}=\int\frac{\text{t}}{\cos^2\text{t}}\text{dt}=\int\text{t}\cdot\sec^2\text{tdt}$
$=\text{t}\cdot\int\sec^2\text{tdt}-\int\Big(\frac{\text{d}}{\text{dt}}\text{t}\cdot\int\sec^2\text{tdt}\Big)\text{dt}$
$=\text{t}\cdot\tan\text{t}-\int1\cdot\tan\text{tdt}$
$=\text{t}\tan\text{t}+\log|\cos\text{t}|+\text{C}$ $\Big[\because\int\tan\text{xdx}=-\log|\cos\text{x}|+\text{C}\Big]$
$=\sin^{-1}\text{x}\cdot\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\log\Big|\sqrt{1-\text{x}^2}\Big|+\text{C}$
View full question & answer→Question 3465 Marks
Evaluate the following integrals:
$\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}(\tan\text{x}+\cot\text{x})^2\text{dx}$
Answer$\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}(\tan\text{x}+\cot\text{x})^2\text{dx}$
$=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\big(\tan^2\text{x}+\cot^2\text{x}+2\tan\text{x }\cot\text{x})\text{dx}$
$=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\big(\sec^2\text{x}-1+\text{cosec}^2\text{x}-1+2\big)\text{dx}$
$=\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\sec^2\text{x dx}+\int_{\frac{\pi}{6}}^\limits{\frac{\pi}{3}}\text{cosec}^2\text{x dx}$
$=\big[\tan\text{x}+(-\cot\text{x})\big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$=\Big(\tan\frac{\pi}{3}-\tan\frac{\pi}{6}\Big)-\Big(\cot\frac{\pi}{3}-\cot\frac{\pi}{6}\Big)$
$=\Big(\sqrt{3}-\frac{1}{\sqrt{3}}\Big)-\Big(\frac{1}{\sqrt{3}}-\sqrt{3}\Big)$
$=2\sqrt{3}-\frac{2}{3}$
$=\frac{4}{\sqrt{3}}$
View full question & answer→Question 3475 Marks
Evaluate the following intregals:
$\int\frac{\text{ax}^2+\text{bx}+\text{c}}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}\ \text{dx},$ where a, b, c are distinct
AnswerWe have
$\text{I}=\int\frac{\text{ax}^2+\text{bx}+\text{c}}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}\ \text{dx}$
Let $\int\frac{\text{ax}^2+\text{bx}+\text{c}}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}=\frac{\text{A}}{\text{x}-\text{a}}+\frac{\text{B}}{(\text{x}-\text{b}) }+\frac{ \text{C}}{(\text{x}-\text{c})}$
$\Rightarrow\text{ax}^2+\text{bx}+\text{c}\\=\text{A}(\text{x}-\text{b})(\text{x}-\text{c})+\text{B}(\text{x}-\text{c})(\text{x}-\text{a})+\text{C}(\text{x}-\text{a})(\text{x}-\text{b}$
$\Rightarrow\text{ax}^2+\text{bx}+\text{c}\\=\text{A}[\text{x}^2-(\text{b}+\text{c})\text{x}+\text{bc}]+\text{B}[\text{x}^2-(\text{c}+\text{a})\text{x}+\text{ca}]\\+\text{C}[\text{x}^2-(\text{a}-\text{b})\text{x}+\text{ab}]$
$\Rightarrow\text{ax}^2+]\text{bx}+\text{c}=(\text{A}+\text{B}+\text{C})\text{x}^2-[\text{A}(\text{b}+\text{c})+\text{B}(\text{c}+\text{a}) \\+\text{C}(\text{a}+\text{b})]\text{x}+\text{Abc}+\text{Bca}+\text{Cab}$
Equation the coefficient on both sides, we get
$\text{a}=\text{A}+\text{B}+\text{C}\ ...(1)$
$\text{b}=-[\text{A}(\text{b}+\text{c})+\text{B}(\text{c}+\text{a})+\text{C}(\text{a}+\text{b})]\ ...(2)$
$\text{c}=\text{Abc}+\text{Bca}+\text{Cab}\ ...(3)$
Solving (1), (2), (3) we get
$\text{A}=\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}$
$\text{B}=\frac{\text{ab}^2+\text{b}^2+\text{c}^2}{(\text{b}-\text{a})(\text{b}-\text{c})}$
$\text{C}=\frac{\text{ac}^2+\text{bc}+\text{c}}{(\text{c}-\text{a})(\text{c}-\text{b})}$
$\therefore\text{I}=\int\Big[\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}\times\frac{1}{\text{x}-\text{a}}+\frac{\text{ab}^2+\text{b}^2+\text{c}}{(\text{b}-\text{a})(\text{b}-\text{c})}\times\frac{1}{\text{x}-\text{b}}\\+\frac{\text{ac}^2+\text{bc}^2+\text{c}}{(\text{c}-\text{a})(\text{c}-\text{b})}\times\frac{1}{\text{x}-\text{c}}\Big]\ \text{dx}$
$=\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}\log|\text{x}-\text{a}|+\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}\log|\text{x}-\text{b}|\\+\frac{\text{ac}^2+\text{bc}^2+\text{c}}{(\text{c}-\text{a})(\text{c}-\text{b})}\log|\text{x}-\text{c}|+\text{K}$
View full question & answer→Question 3485 Marks
Evaluate the following integrals:
$ \int\sqrt{\cot}\theta\text{d}\theta$
Answer$ \int\sqrt{\cot}\theta\text{d}\theta$
Let $\cot\theta=\text{x}^2$
$\Rightarrow-\text{cosec}^2\theta\text{d}\theta=2\text{x dx}$
$\Rightarrow\text{d}\theta=\frac{-2\text{x}}{\text{cosec}^2\theta}\ \text{dx}$
$=\frac{-2\text{x}}{1+\cot^2\theta}$
$=\frac{-2\text{x}}{1+\text{x}^4}\ \text{dx}$
$\therefore\text{I}=-\int\frac{2\text{x}^2}{1+\text{x}^4}\ \text{dx}$
$=-\int\frac{2}{\frac{1}{\text{x}^2}+\text{x}^2}\ \text{dx}$
Dividing numerator and denominator by $x^2$
$=-\frac{1+\frac{1}{\text{x}^2}+1-\frac{1}{\text{x}^2}}{\text{x}^2+\frac{1}{\text{x}^2}}\ \text{dx}$
$=-\int\frac{\Big(1+\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}-\frac{1}{\text{x}^2}\Big)+2}-\int\frac{\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}}{\Big(\text{x}+\frac{1}{\text{x}^2}\Big)^2-2}$
Let $\text{x}-\frac{1}{\text{x}}=\text{t}\Rightarrow\Big(1+\frac{1}{\text{x}^2}\Big)\ \text{dx}=\text{dt}$
and $\text{x}+\frac{1}{\text{x}}=\text{z}\Rightarrow\Big(1-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{dz}$
$\Rightarrow\text{I}=-\int\frac{\text{dt}}{\text{t}^2+2}-\int\frac{\text{dz}}{\text{z}^2-2}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{t}}{\sqrt{2}}\Big)-\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{z}-\sqrt{2}}{\text{z}+\sqrt{2}}\Big|+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}^2-1}{\sqrt{2}\text{x}}\Big)-\frac{1}{2\sqrt{2}}\log\Big|\frac{\text{x}^2+1-\sqrt{2}\text{x}}{\text{x}^2+1+\sqrt{2}\text{x}}\Big|+\text{C}$
$\text{I}=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\cot\theta-1}{\sqrt{2\cot\theta}}\Big)-\frac{1}{2\sqrt{2}}\log\Big|\frac{\cot\theta+1-\sqrt{2\cot\theta}}{\cot\theta+1-\sqrt{2\cot\theta}}\Big|+\text{C}$
View full question & answer→Question 3495 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^3}\sin(\log\text{x})\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}^3}\sin(\log\text{x})\text{dx}$
Putting log x = t
$\Rightarrow\text{x}=\text{e}^\text{t}$
$\Rightarrow\text{dx}=\text{e}^\text{t}\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{\text{e}^{3\text{t}}}\sin\text{t e}^\text{t}\text{dt}$
$=\int\text{e}^{-2\text{t}}\sin\text{t dt}$
Considering sin t as first function and $e^{-2t}$ as second function
$\text{I}=\sin\text{t}\Big[\frac{\text{e}^{-2\text{t}}}{-2}\Big]-\int\cos\text{t}\frac{\text{e}^{-2\text{t}}}{-2}\text{dt}$
$\Rightarrow\text{I}=\frac{\sin\text{t e}^{-2\text{t}}}{-2}+\frac{1}{2}\int\cos\text{t e}^{-2\text{t}}\text{dt}$
$\Rightarrow\text{I}=\frac{\sin\text{t e}^{-2\text{t}}}{-2}+\frac{1}{2}\Big[\cos\text{t}\frac{\text{e}^{-2\text{t}}}{-2}-\int(-\sin\text{t})\frac{\text{e}^{-2\text{t}}}{-2}\text{dt}\Big]$
$\Rightarrow\text{I}=\frac{\sin\text{t e}^{-2\text{t}}}{-2}-\frac{1}{4}\cos\text{t e}^{-2\text{t}}-\int\frac{\text{e}^{-2\text{t}}\sin\text{t dt}}{4}$
$\Rightarrow\text{I}=\text{e}^{-2\text{t}}\Big[\frac{-2\sin\text{t}-\cos\text{t}}{4}\Big]-\frac{\text{I}}{4}$
$\Rightarrow\frac{5\text{I}}{4}=\text{e}^{-2\text{t}}\Big[\frac{-2\sin\text{t}-\cos\text{t}}{4}\Big]$
$\Rightarrow\text{I}=\frac{\text{e}^{-2\text{t}}}{5}[-2\sin\text{t}-\cos\text{t}]+\text{C}$
$\Rightarrow\text{I}=\frac{-\text{x}^{-2}}{5}[2\sin(\log\text{x})+\cos(\log\text{x})]+\text{C}$
$\Rightarrow\text{I}=\frac{-1}{5\text{x}^2}[\cos(\log\text{x})+2\sin(\log\text{x})]+\text{C}$
View full question & answer→Question 3505 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}\tan^{-1}\text{x})}{(1+\text{x}^2)^{\frac{3}{2}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}\tan^{-1}\text{x}}{(1+\text{x}^2)^{\frac{3}{2}}}\text{dx}$
putting $\text{x}=\tan\theta$
$\Rightarrow\text{dx}=\sec^2 \theta\text{d}\theta$
$\&\theta\tan^{-1}\text{x}$
$\therefore\text{I}=\int\frac{(\tan\theta).\theta\sec^2\theta\text{d}\theta}{\big(1+\tan^2\theta\big)^{\frac{3}{2}}}$
$=\int\frac{\theta.\tan\theta\sec^2\theta\text{d}\theta}{(\sec^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\theta\tan\theta.\sec^2\theta\text{d}\theta}{\sec^3\theta}$
$=\int\frac{\theta.\tan\theta}{\sec\theta}\text{d}\theta$
$=\int\theta.\sin\theta\text{d}\theta$
$=\theta\int\sin\theta\text{d}\theta-\int\big\{\frac{\text{d}}{\text{d}\theta}(\theta)\int\sin\text{d}\theta\big\}\text{d}\theta$
$=\theta(-\cos\theta)-\int1.(-\cos\theta)\text{d}\theta$
$=-\theta\cos\theta+\sin\theta+\text{C}$
$=\frac{-\theta}{\sec\theta}+\frac{1}{\text{cosec}\theta}+\text{C}$
$=\frac{-\theta}{\sqrt{1+\tan^2\theta}}+\frac{1}{\sqrt{1+\cot^2}\theta}+\text{C}$
$=\frac{-\theta}{\sqrt{1+\tan^2\theta}}+\frac{\tan\theta}{\sqrt{\tan^2\theta+1}}+\text{C}$
$=\frac{-\tan^{-1}\text{x}}{\sqrt{1+\text{x}^2}}+\frac{\text{x}}{\sqrt{\text{x}^2+1}}+\text{C}$
View full question & answer→