Question 512 Marks
Write a value of $\int\tan\text{x}\sec^3\text{x dx}$
AnswerLet $\text{I}=\int\tan\text{x}\sec^3\text{x dx}$
Let $\sec\text{x}=\text{t}$
$\sec\text{x}\tan\text{x dx}=\text{dt}$
$\text{dx}=\frac{\text{dt}}{\sec\text{x}\tan\text{x}}$
$\therefore\ \text{I}=\int\sec^2\text{x}\tan\text{x dx}$
$=\int\text{t}^2\text{ dt}$
$=\frac{\text{t}^3}{3}+\text{C}$
$=\frac{\sec^3\text{x}}{3}+\text{C}$
$\therefore\ \text{I}=\frac{\sec^3\text{x}}{3}+\text{C}$
View full question & answer→Question 522 Marks
Integrate the function in Exercise.
$\sqrt{4-\text{x}^2}$
Answer$\int\sqrt{4-\text{x}^2}\text{dx}=\int\sqrt{2^2-\text{x}^2}\text{dx}$
$=\frac{\text{x}}{2}\sqrt{2^2-\text{x}^2}+\frac{2^2}{2}\sin^{-1}\frac{\text{x}}{2}+\text{C}$
$\Bigg[\therefore\ \sqrt{\text{a}^2-\text{x}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}\Bigg]$
$=\frac{\text{x}}{2}\sqrt{4-\text{x}^2}+2\sin^{-1}\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 532 Marks
By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\frac{\sin\text{x}-\cos\text{x}}{1+\sin\text{x}\cos\text{x}}\text{dx}$
Answer$\text{Let}\text{I}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\sin\text{x}-\cos\text{x}}{1+\sin\text{x}\cos\text{x}}\text{dx}$
$\Rightarrow\ \ \text{I}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)-\cos\bigg(\frac{\pi}{2}-\text{x}\bigg)}{1+\sin\bigg(\frac{\pi}{2}-\text{x}\bigg)\cos\bigg(\frac{\pi}{2}-\text{x}\bigg)}\text{dx}=\int^{\frac{\pi}{2}}\limits_{0}\frac{\cos\text{x}-\sin\text{x}}{1+\cos\text{x}\sin\text{x}}\text{dx}=-\int^{\frac{\pi}{2}}\limits_{0}\frac{\sin\text{x}-\cos\text{x}}{1+\sin\text{x}\cos\text{x}}\text{dx}$
Adding eq. (i) and (ii), we have $21=0 \ \ \Rightarrow 1=0$
View full question & answer→Question 542 Marks
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\sin^{2}\text{x dx}$
Answer$\int\limits^\frac{\pi}{2}_0\sin^2\text{x}\text{dx}$
$= \int\limits^\frac{\pi}{2}_0\frac{1-\cos2\text{x}}{2}\text{dx}$
$= \frac{1}{2}\int\limits^\frac{\pi}{2}_0(1-\cos2\text{x})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{x}}{2}\Big]^\frac{\pi}{2}_0$
$=\frac{1}{2}(\frac{\pi}{2}-0)$
$=\frac{\pi}{4}$
View full question & answer→Question 552 Marks
If f is an integrable function, show that:
$\int\limits^{\text{a}}_{-\text{a}}\text{xf}\big(\text{x}^2\big)\text{dx}=0$
Answer$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{xf}(\text{x}^2)\text{dx}$
Let $\text{g(x)}=\text{f}\big(\text{x}^2\big)$
$\Rightarrow\text{g}(-\text{x})=(-\text{x})\text{f}(-\text{x})^2=-(\text{x})\text{f}(\text{x})^2=-\text{g(x)}\text{ i.e., g(x) is even}$
Therefore,
$\text{I}=0$ $\bigg[\text{Using}\int\limits^{\text{a}}_{-\text{a}}\text{g}(\text{x})\text{dx}=0\text{ when g(x) is odd}\bigg]$
View full question & answer→Question 562 Marks
Integrate the function in Exercise:$\text{e}^{3}\log\text{x}\big(\text{x}^{4}+1\big)^{-1}$
Answer$ \text{e}^{3\log\text{x}}.\big(\text{x}^{4}+1\big)^{-1}=\text{e}^{\log\text{x}^{3}}.\big(\text{x}^{4}+1\big)^{-1}=\frac{\text{x}^{3}}{\big(\text{x}^{4}+1\big)}$
$\text{Let}\ \text{x}^{4}+1=\text{t}\Rightarrow 4\text{x}^3\ \text{dx}=\text{dt}$
$\Rightarrow\int\text{e}^{3\log\text{x}}\big(\text{x}^{4}+1\big)^{-1}\text{dx}=\int\frac{\text{x}^{3}}{\big(\text{x}^{4}+1\big)}\text{dx}$
$=\frac{1}{4}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{4}\log|\text{t}|+\text{C}$
$=\frac{1}{4}\log\big(\text{x}^{4}+1\big)+\text{C}$
View full question & answer→Question 572 Marks
By Using properties of definite integral, evaluate the following integral in Exercise:
$\int^{\frac{\pi}{2}}_{\frac{-\pi}{2}}\sin^{7}\text{x}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\sin^{7}\text{x}\ \text{dx}$
$\text{Here}\ \ \text{f}\text{(x)}=\sin^{7}\text{x}$
$\therefore\ \ \text{f}(-\text{x)}=\sin^{7}(-\text{x})=(-\sin^{7}\text{x})=-\sin^{7}\text{x}=-\text{f}\text{(x)}$
$\therefore\ \ \text{f}(\text{x})\ \text{is an odd function of x.}$
$\therefore\ \ \text{I}=\int^{\frac{\pi}{2}}\limits_{\frac{-\pi}{2}}\sin^{7}\text{x}\ \text{dx}=0\ \ \bigg[\because\int^{\text{a}}\limits_{-\text{a}}\text{f}\text{(x)}\text{dx}=0,$when $\text{f(x)}$ is an odd function$\bigg]$
View full question & answer→Question 582 Marks
Evaluate $\int\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Let $\sqrt{\text{x}}=\text{t}$
$\frac{1}{2\sqrt{\text{x}}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=2\int\sin\text{t dt}$
$=-2\cos\text{t}+\text{C}$
$\text{I}=-2\cos\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 592 Marks
Evaluate $\int\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}}$
Let $\text{x}=\sin\theta$
$\text{dx}=\cos\theta$
$\therefore\ \text{I}=\int\frac{\cos\theta}{\cos\theta}\text{ d}\theta$
$=\int\text{d}\theta$
$=\theta+\text{C}$
$=\sin^{-1}\text{x}+\text{C}$ $(\because\text{ x}=\sin\theta)$
View full question & answer→Question 602 Marks
Evaluate the following integrals:
$\int\limits^2_02\text{x}\big[\text{x}\big]\text{dx}$
Answer[x] = 0 for 0
and [x] = 1 for 1
Hence,
$=\int\limits^1_10+\int\limits^2_12\text{x}\text{ dx}$
$=\big\{\text{x}^2\big\}^2_1$
$=3$
View full question & answer→Question 612 Marks
Write a value of $\int\sin^3\text{x}\cos\text{x dx}$
AnswerLet $\text{I}=\int\sin^3\text{x}\cos\text{x dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{t}^3\text{ dt}$
$=\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\sin^4\text{x}}{4}+\text{C}$ $(\because\text{t}=\sin\text{x})$
View full question & answer→Question 622 Marks
Evaluate the integral in Exercise:
$\int_{0}^{1}\frac{\text{x}}{\text{x}^{2}+1}\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{1}\frac{\text{x}}{\text{x}^{2}+1}\text{dx}$
$\text{put}\ \text{x}^{2}+1=\text{y},\ \ \therefore 2\text{x}\ \text{dx}=\text{dy},\ \text{or}\ \text{x}\ \text{dx}=\frac{1}{2}\text{dy}$
$\text{when}\ \text{x}=2,\ \text{y}=5$
$\text{when}\ \text{x}=3,\ \text{y}=10$
$\therefore |=\frac{1}{2}\int^{10}_{5}\frac{\text{dy}}{\text{y}}=\frac{1}{2}\big[\log\text{y}\big]^{10}_{5}=\frac{1}{2}\big[\log10-\log5\big]=\frac{1}{2}\log\bigg[\frac{10}{5}\bigg]=\frac{1}{2}\log2$
View full question & answer→Question 632 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{\text{x}}(\sqrt{\text{x}}+1)}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{\text{x}}(\sqrt{\text{x}}+1)}\text{dx}$
Putting $\sqrt{\text{x}}+1=\text{t}$
$\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{\text{x}}}\text{dx}=2\text{dt}$
$\therefore\text{I}=2\int\frac{1}{\text{t}}\text{dt}$
$=2\text{ In }|\text{t}|+\text{C}$
$=2\text{ In }|\sqrt{\text{x}}+1|+\text{C }\big[\because\text{t}=\sqrt{\text{x}}+1\big]$
View full question & answer→Question 642 Marks
Evalute the following integrals:
$\int\frac{\text{cosec}^2\text{x}}{1+\cot\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{cosec}^2\text{x}}{1+\cot\text{x}}\text{dx}$
Putting $\cot\text{x}=\text{t}$
$\Rightarrow-\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{cosec}^2\text{x dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{dt}}{1+\text{t}}$
$=-\text{ln}|1+\text{t}|+\text{C}$
$=-\text{ln}|1+\cot\text{x}|+\text{C}\ \big[\because\text{t}=\cot\text{x}\big]$
View full question & answer→Question 652 Marks
Evalute the following integrals:
$\int\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\text{dx}$
Answer$\int\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\text{dx}$
$=\int\sqrt{\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}}\text{dx}$
$\Big[\because 1-\cos\text{x}=2\sin^2\frac{\text{x}}{2} \ \&\ 1+\cos\text{x}=2\cos^2\frac{\text{x}}{2}\Big]$
$=\int\tan\frac{\text{x}}{2}\text{dx}$
$=-2\text{ln}\Big|\cos\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 662 Marks
Integrate the function in Exercise:
$\text{e}^\text{x}(\sin\text{x}+\cos\text{x})$
AnswerLet $\text{I}=\int\text{e}^\text{x}(\sin\text{x}+\cos\text{x})\text{dx}$
Let $\text{f}(\text{x})=\sin\text{x}$
$\Rightarrow \ \text{f}'(\text{x})=\cos\text{x}$
$\therefore\ \text{I}=\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}$
It is known that, $=\int\text{e}^\text{x}\{\text{f}(\text{x})+\text{f}'(\text{x})\}\text{dx}=\text{e}^\text{x}\text{f}(\text{x})+\text{C}$
$\therefore\ \text{I}=\text{e}^\text{x}\sin\text{x}+\text{C}$
View full question & answer→Question 672 Marks
Find the integrals of the functions in Exercises:
$\frac{\sin^3\text{x}+\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}$
Answer$\frac{\sin^3\text{x}+\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}$
$=\frac{\sin^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}+\frac{\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}$
$=\frac{\sin\text{x}}{\cos^2\text{x}}+\frac{\cos\text{x}}{\sin^2\text{x}}$
$=\tan\text{x}\sec\text{x}+\cot\text{x}\text{ cosec x}$
$\therefore\ \ \ \int\frac{\sin^3\text{x}+\cos^3\text{x}}{\sin^2\text{x}\cos^2\text{x}}\text{ dx}=\int\big(\tan\text{x}\sec\text{x}+\cot\text{x}\text{ cosec x}\big)\text{dx}$
$=\sec\text{x}-\text{cosec x}+\text{C}$
View full question & answer→Question 682 Marks
$\int\tan^2(2\text{x}-3)\text{dx}$
Answer$\int\tan^2(2\text{x}-3)\text{dx}$
$=\int[\sec^2(2\text{x}-3)-1]\text{dx}$
$=\int\sec^2(2\text{x}-3)\text{dx}-\int1\text{dx}$
$=\frac{\tan(2\text{x}-3)}{2}-\text{x}+\text{c}$
View full question & answer→Question 692 Marks
Evaluate the following integrals:
$\int\Big\{3\sin\text{x}-4\cos\text{x}+\frac{5}{\cos^2\text{x}}-\frac{6}{\sin^2\text{x}}+\tan^2\text{x}-\cot^2\text{x}\Big\}\text{dx}$
Answer$\int\Big\{3\sin\text{x}-4\cos\text{x}+\frac{5}{\cos^2\text{x}}-\frac{6}{\sin^2\text{x}}+\tan^2\text{x}-\cot^2\text{x}\Big\}\text{dx}$
$=3\int\sin\text{x dx}-4\int\cos\text{x dx}+5\int\sec^2\text{dx}\\-6\int\text{cosec}^2\text{x}+\int\tan^2\text{x dx}-\int\cot^2\text{x dx}$
$=3\int\sin\text{x dx}-4\int\cos\text{x dx}+5\int\sec^2\text{x dx}\\-6\int\text{cosec}^2\text{x}+\int(\sec^2\text{x}-1)\text{dx}-\int(\text{cosec}^2\text{x}-1)\text{dx}$
$=3\int\sin\text{x dx}-4\int\cos\text{x dx}+6\int\sec^2\text{x dx}-7\int\text{cosec}^2\text{x dx}$
$=-3\cos\text{x}-4\sin\text{x}+6\tan\text{x}+7\cot\text{x}+\text{C}$
View full question & answer→Question 702 Marks
$\int\sin^2\frac{\text{x}}{2}\text{dx}$
AnswerLet I $=\int\sin^2\frac{\text{x}}{2}\text{dx}.$ Then,
$\text{I}=\frac{1}{2}\int2\sin^2\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\int(1-\cos\text{x})\text{dx}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$
$=\frac{1}{2}\int\text{dx}-\frac{1}{2}\int\cos\text{xdx}$
$=\frac{1}{2}\times\text{x}-\frac{1}{2}\times\sin\text{x}+\text{C}$
$=\frac{1}{2}(\text{x}-\sin\text{x})+\text{C}$
$\therefore\text{I}=\frac{1}{2}(\text{x}-\sin\text{x})+\text{C}$
View full question & answer→Question 712 Marks
Evaluate the definite integral in Exercise:
$\int\limits_{0}^{1}\frac{\text{dx}}{\sqrt{1-\text{x}^{2}}}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{1}\frac{\text{dx}}{\sqrt{1-\text{x}^{2}}}$ $\int\frac{\text{dx}}{\sqrt{1-\text{x}^{2}}}=\sin^{-1}\text{x}=\text{F}\text{(x)}$ By second fundamental theorem of calculus, we obtain $\text{I}=\text{F}(1)-\text{F}(0)$ $=\sin^{-1}(1)-\sin^{-1}(0)$ $=\frac{\pi}{2}-0$$=\frac{\pi}{2}$
View full question & answer→Question 722 Marks
Evaluate the following integrals:
$\int\limits^2_0\sqrt{4-\text{x}^2}\text{ dx}$
Answer$\int\limits^2_0\sqrt{4-\text{x}^2}\text{ dx}$
$=\int\limits^2_0\sqrt{2^2-\text{x}^2}\text{ dx}$
$=\Big[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}\text{ dx}+\frac{1}{2}\times2^2\sin^{-1}\frac{\text{x}}{2}\Big]^2_0$
$=\Big[\frac{\text{x}}{2}\sqrt{4-\text{x}^2}\text{ dx}\Big]^2_0+2\Big[\sin^{-1}\frac{\text{x}}{2}\Big]^2_0$
$=0+2\Big(\frac{\pi}{2}-0\Big)$
$=\pi$
View full question & answer→Question 732 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{1}_{0}\text{x}(1-\text{x})^{\text{n}}\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{1}\text{x}(1-\text{x})^{\text{n}}\ \text{dx}=\int\limits_{0}^{1}(1-\text{x)}\left\{1-(1-\text{x)}^{\text{n}}\right\}\text{dx}\ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}\text{(a}-\text{x})\text{dx}=\bigg]$
$\Rightarrow\ \text{I}=\int\limits_{0}^{1}(1-\text{x})(1-1+\text{x})^{\text{n}}\ \text{dx}=\int\limits_{0}^{1}(1-\text{x})\text{x}^{\text{n}}\ \text{dx}=\int^{1}_{0}(\text{x}^{\text{n}}-\text{x}^{\text{n}+1})\text{dx}$
$\Rightarrow\ \ \text{I}=\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}-\frac{\text{x}^{\text{n}+2}}{\text{n}+2}\bigg)^{1}_{0}=\frac{1}{\text{n}+1}-\frac{1}{\text{n}+2}-(0-0)=\frac{\text{n}+2-\text{n}-1}{(\text{n}+1)(\text{n}+2)}=\frac{1}{\text{(n}+1)(\text{n}+2)}$
View full question & answer→Question 742 Marks
If f is an integrable function, show that:
$\int\limits^{\text{a}}_{-\text{a}}\text{f}\big(\text{x}^2\big)\text{dx}=2\int\limits^\text{a}_0\text{f}\big(\text{x}^2\big)\text{dx}$
Answer$\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{f}\big(\text{x}^2\big)\text{dx}$
Here, $\text{g(x)}=\text{f}\big(\text{x}^2\big)$
$\Rightarrow\text{g}(-\text{x})=\text{f}(-\text{x})^2=\text{f}(\text{x})^2=\text{g(x)}\text{ i.e., g(x) is even}$
Therefore,
$\text{I}=2\int\limits^{\text{a}}_{0}\text{f}(\text{x}^2)\text{dx}$ $\bigg[\text{Using}\text{I}=\int\limits^{\text{a}}_{-\text{a}}\text{g}(\text{x})\text{dx}=\int\limits^{\text{a}}_{0}\text{f}(\text{x})\text{dx}\text{ when g(x) is even}\bigg]$
View full question & answer→Question 752 Marks
Evalute the following integrals:
$\int\sqrt{\frac{1+\cos2\text{x}}{1-\cos2\text{x}}}\text{dx}$
Answer$\int\sqrt{\frac{1+\cos2\text{x}}{1-\cos2\text{x}}}\text{dx}$
$=\int\sqrt{\frac{2\cos^2\text{x}}{2\sin^2\text{x}}}\text{dx}$
$=\int\cot\text{x dx}$
$=\text{ln}|\sin\text{x}|+\text{C}$
View full question & answer→Question 762 Marks
Evaluate the following integrals:$\int\text{x}^{\text{n}}.\log\text{x dx}$
Answer$\int\text{x}^{\text{n}}\log\text{x dx}$
Taking $\log x$ as the first function and $x^n$ as the second function.
$=\log\text{x}\int\text{x}^\text{n}\text{dx}-\int\Big(\frac{\text{d}}{\text{dx}}\log\text{x}\int\text{x}^\text{n}\text{dx}\Big)\text{dx}$
$=\log\text{x}\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}\bigg)-\int\frac{1}{\text{x}}\bigg(\frac{\text{x}^{n+1}}{\text{n}+1}\bigg)\text{dx}$
$=\log\text{x}\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}\bigg)-\int\frac{\text{x}^{\text{n}}}{\text{n}+1}\text{dx}$
$=\log\text{x}\bigg(\frac{\text{x}^{\text{n}+1}}{\text{n}+1}\bigg)-\int\frac{\text{x}^{\text{n}+1}}{(\text{n}+1)^2}+\text{C}$
View full question & answer→Question 772 Marks
Write the coefficient a, b, c of which the value of the integral $\int\limits^3_{-3}(\text{ax}^2+\text{bx}+\text{c})\text{dx}$ is independent.
Answer$\int\limits^3_{-3}(\text{ax}^2+\text{bx}+\text{c})\text{dx}$
$=\Big[\text{a}\frac{\text{x}^3}{3}+\text{b}\frac{\text{x}^2}{2}+\text{cx}\Big]^3_{-3}$
$=9\text{a}+\frac{9}{2}\text{b}+3\text{c}+9\text{a}-\frac{9}{2}\text{b}+3\text{c}$
$=18\text{a}+6\text{c}$
Hence, the given integral is independent of b.
View full question & answer→Question 782 Marks
Evaluate the following integrals:
$\int\limits^{4}_2\frac{\text{x}}{\text{x}^2+1}\text{dx}$
Answer$\int\limits^{4}_2\frac{\text{x}}{\text{x}^2+1}\text{dx}$
$=\frac{1}{2}\int\limits^{4}_2\frac{2\text{x}}{\text{x}^2+1}\text{dx}$
$=\frac{1}{2}\times\Big[\log(\text{x})^2-1\Big]^4_2$ $\Big[\int\frac{\text{f}'(\text{x})}{\text{f(x)}}\text{ dx}=\log\text{f(x)}+\text{C}\Big]$
$=\frac{1}{2}\big(\log17-\log5\big)$
$=\frac{1}{2}\log\Big(\frac{17}{5}\Big)$ $\Big(\log\text{a}-\log\text{b}=\log\frac{\text{a}}{\text{b}}\Big)$
View full question & answer→Question 792 Marks
Evaluate the following:
$\int\frac{\text{dt}}{\sqrt{3\text{t}-2\text{t}^2}}$
AnswerLet $\text{I}=\int\frac{\text{dt}}{\sqrt{3\text{t}-2\text{t}^2}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{-\Big(\text{t}^2-\frac{3}{2}\text{t}\Big)}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\Big(\frac{3}{4}\Big)^2-\Big(\text{t}-\frac{3}{4}\Big)^2}}$
$=\frac{1}{\sqrt{2}}\sin^{-1}\bigg(\frac{\text{t}-\frac{3}{4}}{\frac{3}{4}}\bigg)+\text{C}$ $=\frac{1}{\sqrt{2}}\sin^{-1}\Big(\frac{4\text{t}-3}{3}\Big)+\text{C}$
View full question & answer→Question 802 Marks
Evaluate the following integrals:
$\int\limits^{1}_02^{\text{x}-[\text{x}]}\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{1}_02^{\text{x}-[\text{x}]}\text{dx}$
$=\int\limits^{1}_02^{\text{x}-0}\text{ dx}$ $\big(\because[\text{x}]=0,\text{ where}, 0<\text{x}<1\big)$
$=\int\limits^{1}_02^{\text{x}}\text{ dx}$
$=\Big[\frac{2^{\text{x}}}{\log_\text{e}2}\Big]^1_0$
$=\frac{2^1}{\log_\text{e}2}-\frac{2^0}{\log_\text{e}2}$
$=\frac{2}{\log_\text{e}2}-\frac{1}{\log_\text{e}2}$
$=\frac{1}{\log_\text{e}2}$
View full question & answer→Question 812 Marks
Evaluate the following integrals:
$\int\text{e}^{\cos^2\text{x}}\sin2\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\text{e}^{\cos^2\text{x}}\sin2\text{x}\text{ dx}$
Let $\cos^2\text{x}=\text{t}$
On differentiating both sides, we get
$-2\cos\text{x}\sin\text{x}\text{ dx}=\text{dt}$
$\therefore\text{I}=\int\text{e}^\text{t}2\sin\text{x}\cos\text{x}\frac{\text{dt}}{-2\sin\text{x}\cos\text{x}}$
$=-\int\text{e}^\text{t}\text{dt}$
$=-\text{e}^\text{t}+\text{C}$
$=-\text{e}^{\cos^2\text{x}}+\text{C}$
View full question & answer→Question 822 Marks
If $\int_{0}^\limits{\text{a}}3\text{x}^2\text{ dx}=8,$ find the value of a.
AnswerWe have,
$\int_{0}^\limits{\text{a}}3\text{x}^2\text{ dx}=8$
$\Rightarrow\Big[3-\frac{\text{x}^3}{3}\Big]^{\text{a}}_0=8$
$\Rightarrow\text{a}^3=8$
$\Rightarrow\text{a}=2$
View full question & answer→Question 832 Marks
Evaluate the following integrals:
$\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$ Let $\sqrt{\text{x}}=\text{t}$ $\Rightarrow\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\sqrt{\text{x}}}=2\text{dt}$Now, $\int\frac{\cos\sqrt{\text{x}}}{\sqrt{\text{x}}}\text{ dx}$
$=2\int\cos\text{t dt}$
$=2\sin\text{t}+\text{C}$
$=2\sin\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 842 Marks
Integrate the functions in Exercises:
$\frac{1}{\sqrt{9-25\text{x}^2}}$
Answer$\int\frac{1}{\sqrt{9-25\text{x}^2}}\text{ dx}=\int\frac{1}{\sqrt{(3)^2-(5\text{x})^2}}\text{ dx}$
$=\frac{\sin^{-1}\frac{5\text{x}}{3}}{5\rightarrow\text{Coeff. of x}}+\text{c}$$\ \ \ \ \ \ \ \bigg[\because\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{ dx}=\sin^{-1}\frac{\text{x}}{\text{a}}\bigg]$
$=\frac{1}{5}\sin^{-1}\bigg(\frac{5\text{x}}{3}\bigg)+\text{c}$
View full question & answer→Question 852 Marks
Evaluate:
$\int\sqrt{\frac{1-\cos\ 2\text{x}}{2}}\text{dx}$
Answer$\int\sqrt{\frac{1-\cos\ 2\text{x}}{2}}\text{dx}$
$\int\sqrt{\frac{2\sin^2\text{x}}{2}}\text{dx}\ \ [\therefore1-\cos2\text{x}=2\sin^2\text{x]}$
$=\int\sin\text{x dx}$
$=-\cos\text{x}+\text{c}$
View full question & answer→Question 862 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\sqrt{\text{x}}}\cos\big(\text{e}^{\sqrt{\text{x}}}\big)}{\sqrt{\text{x}}}\text{ dx}$
Answer$\int\frac{\text{e}^{\sqrt{\text{x}}}\cos\big(\text{e}^{\sqrt{\text{x}}}\big)}{\sqrt{\text{x}}}\text{ dx}$
Let $\text{e}^{\sqrt{\text{x}}}=\text{t}$
$\Rightarrow\text{e}^{\sqrt{\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{e}^{\sqrt{\text{x}}}}{\sqrt{\text{x}}}\text{ dx}=2\text{dt}$
Now, $\int\frac{\text{e}^{\sqrt{\text{x}}}\cos\big(\text{e}^{\sqrt{\text{x}}}\big)}{\sqrt{\text{x}}}\text{ dx}$
$=2\int\cos\text{t}\text{ dt}$
$=2\sin\text{t}+\text{C}$
$=2\sin\Big(\text{e}^\sqrt{\text{x}}\Big)+\text{C}$
View full question & answer→Question 872 Marks
Evaluate $\int\frac{2}{1-\cos2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2}{1-\cos2\text{x}}\text{ dx}$
$=\int\frac{2}{\sin^2\text{x}+\cos^2\text{x}-(\cos^2\text{x}-\sin^2\text{x})}\text{ dx}$
$=\int\frac{2}{2\sin^2\text{x}}\text{ dx}$
$=\int\frac{1}{\sin^2\text{x}}\text{ dx}$
$=\int\text{cosec}^2\text{x}\text{ dx}$
$=-\cot\text{x}+\text{C}$
View full question & answer→Question 882 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\cos2\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\cos(2\text{x})}\text{dx}$ $\Big[\therefore\ 1-\cos\text{A}=2\sin^2\Big(\frac{\text{A}}{2}\Big)\Big]$
$=\int\frac{\text{dx}}{2\sin^2\text{x}}$
$=\frac{1}{2}\int\text{cosec}^2\text{x dx}$
$=\frac{1}{2}[-\cot\text{x}]+\text{C}$
$=-\frac{1}{2}\cot\text{x}+\text{C}$
View full question & answer→Question 892 Marks
Evalute the following integrals:
$\int\frac{1-\sin\text{x}}{\text{x}+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\sin\text{x}}{\text{x}+\cos\text{x}}\text{dx}$
Putting $\text{x}+\cos\text{x}=\text{t}$
$\Rightarrow1-\sin\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(1-\sin\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln t}+\text{C}$
$=\text{ln}|\text{x}+\cos\text{x}|+\text{C}\ \big[\because\text{t}=\text{x}+\cos\text{x}\big]$
View full question & answer→Question 902 Marks
Evaluate the following integrals:
$\int\cos^{-1}(\sin\text{x})\text{dx}$
Answer$\int\cos^{-1}(\sin\text{x})\text{dx}$
$=\int\cos^{-1}\Big(\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big)\text{dx}$ $\Big[\therefore\ \sin\text{x}=\cos\Big(\frac{\pi}{2}-\text{x}\Big)\Big]$
$=\int\Big(\frac{\pi}{2}-\text{x}\Big)\text{dx}$
$=\frac{\pi\text{x}}{2}-\frac{\text{x}^2}{2}+\text{C}$
View full question & answer→Question 912 Marks
Evaluate $\int\frac{\text{e}^{\tan^{-1\text{x}}}}{1+\text{x}^2}\text{ dx}$
AnswerLet $\tan^{-1}\text{x}=\text{t}$
$\frac{1}{1+\text{x}^2}\text{ dx}=\text{dt}$
Let $\text{I}=\int\frac{\text{e}^{\tan^{-1\text{x}}}}{1+\text{x}^2}\text{ dx}$
$=\int\text{e}^{\text{t}}\text{dt}$
$=\text{e}^{\text{t}}+\text{C}$
$=\text{e}^{\tan^{-1}}\text{x}+\text{C}$
View full question & answer→Question 922 Marks
Evaluate:
$\int\frac{\cos2\text{x}+2\sin^2\text{x}}{\sin^2\text{x}}\text{dx}$
Answer$\int\Big(\frac{\cos2\text{x}+2\sin^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1-2\sin^2\text{x}+2\sin^2\text{x}}{\sin^2\text{x}}\Big)\text{dx}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$
$=\int\text{cosec}^2\text{x}\text{ dx}$
$=-\cot\text{x}+\text{c}$
View full question & answer→Question 932 Marks
Evaluate the following integrals:
$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$
Answer$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$
$=\int\limits^{\text{e}^2}_\text{e}\frac{\frac{1}{\text{x}}}{\log\text{x}}\text{ dx}$
$=\log\big[(\log{\text{x}})\big]^{\text{e}^2}_\text{e}$ $\Big[\int\frac{\text{f}'(\text{x})}{\text{f(x)}}\text{ dx}=\log\text{f(x)}+\text{C}\Big]$
$=\log\big(\log\text{e}^2\big)-\log(\log\text{e})$
$=\log(2\log\text{e})-\log(\log\text{e})$
$=\log2-\log1$ $(\log\text{e}=1)$
$=\log2-0$
$=\log2$
View full question & answer→Question 942 Marks
Verify the following:
$\int\frac{2\text{x}+3}{\text{x}^2+3\text{x}}\text{dx}=\log\big|\text{x}^2+3\text{x}\big|+\text{c}$
AnswerLet $\text{I}=\int\frac{2\text{x}+3}{\text{x}^2+3\text{x}}\text{dx}$
Put $\text{x}^2+3\text{x}=\text{t}$
$\Rightarrow\ (2\text{x}+3)\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{1}{\text{t}}\text{dt}=\log|\text{t}|+\text{C}$ $=\log\big|(\text{x}^2+3\text{x})\big|+\text{C}$
View full question & answer→Question 952 Marks
Integrate the function in Exercise:
$\text{x} \ \sec^2\text{x}$
AnswerLet $\text{I}=\int\text{x}\sec^2\text{x dx}$
Taking $x$ as first function and $\sec^2x$ as second function and integrating by parts, we obtain.
$\text{I}=\text{x}\int\sec^2\text{x dx}-\int\Big[\Big\{\frac{\text{d}}{\text{dx}}\text{x}\int\sec^2\text{x} \ \text{dx}\Big\}\text{dx}\Big]$
$=\text{x}\tan\text{x}-\int1.\tan\text{x dx}$
$=\text{x}\tan\text{x}+\text{log}|\cos\text{x}|+\text{C}$
View full question & answer→Question 962 Marks
$\int\sin^2\text{bx dx}$
Answer$\int\sin^2\text{bx dx}$
$=\int\Big[\frac{1-\cos2\text{bx}}{2}\Big]\text{dx}$ $\Big[\therefore\sin^2\text{x}=\frac{1-\cos2\text{x}}{2}\Big]$
$=\frac{1}{2}\int(1-\cos2\text{bx})\text{dx}$
$=\frac{1}{2}\Big[\text{x}-\frac{\sin2\text{bx}}{2\text{b}}\Big]+\text{C}$
$=\frac{\text{x}}{2}-\frac{\sin2\text{bx}}{4\text{b}}+\text{C}$
View full question & answer→Question 972 Marks
Evalute the following integrals:
$\int\frac{\text{x}+1}{\text{x}(\text{x}+\log\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\text{x}+1}{\text{x}(\text{x}+\log\text{x})}\text{dx}$
Putting $\text{x}+\log\text{x}=\text{t}$
$\Rightarrow1+\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{x}+1}{\text{x}}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\text{x}+\log\text{x}|+\text{C}$
View full question & answer→Question 982 Marks
Evaluate the following integral:
$\int\frac{1}{\text{a}^2-\text{b}^2\text{x}^2}\text{ dx}$
Answer$\int\frac{1}{\text{a}^2-\text{b}^2\text{x}^2}\text{ dx}$
$=\frac{1}{\text{b}^2}\int\frac{\text{dx}}{\text{a}^2-\text{b}^2\text{x}}$
$=\frac{1}{\text{b}^2}\times\frac{1}{2\frac{\text{a}}{\text{b}}}\log\Bigg|\frac{\frac{\text{a}}{\text{b}}+\text{x}}{\frac{\text{a}}{\text{b}}-\text{x}}\Bigg|+\text{C}$ $\Big[\therefore\int\frac{\text{dx}}{\text{a}^2-\text{x}^2}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{x}}{\text{a}-\text{x}}\Big|+\text{C}\Big]$
$=\frac{1}{2\text{ab}}=\frac{1}{2\text{a}}\log\Big|\frac{\text{a}+\text{bx}}{\text{a}-\text{bx}}\Big|+\text{C}$
View full question & answer→Question 992 Marks
Evaluate the following integral:
$\int\frac{1}{\sqrt{\text{a}^2-\text{b}^2\text{x}^2}}\text{ dx}$
Answer$\int\frac{1}{\sqrt{\text{a}^2-\text{b}^2\text{x}^2}}\text{ dx}$
$=\int\frac{\text{dx}}{\sqrt{\text{b}^2\Big(\frac{\text{a}^2}{\text{b}^2}-\text{x}^2}\Big)}$
$=\frac{1}{\text{b}}\int\frac{\text{dx}}{\sqrt{\big(\frac{\text{a}}{\text{b}}\big)^2-\text{x}^2}}$
$=\frac{1}{\text{b}}\sin^{-1}\Big(\frac{\text{xb}}{\text{a}}\Big)+\text{C}$
View full question & answer→Question 1002 Marks
Integrate the functions in Exercises:
$\frac{3\text{x}^2}{\text{x}^6+1}$
Answer$\text{Let I}=\int\frac{3\text{x}^2}{\text{x}^6+1}\text{ dx}$
$=\int\frac{3\text{x}^2}{(\text{x}^3)^2+1}\text{ dx} \ \ \ \ \ ...\text{(i)}$
Putting$\ \ \ \text{x}^3=\text{t} \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ 3\text{x}^2=\frac{\text{dt}}{\text{dx}} \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ 3\text{x}^2\text{ dx}=\text{dt}$
$\therefore \ \ \ \ \ $From eq. (i),$\ \ \ \ \text{I}=\int\frac{\text{dt}}{\text{t}^2+1}=\frac{1}{1}\tan^{-1}\frac{\text{t}}{1}+\text{c}$
$=\tan^{-1}\text{x}^3+\text{c}$
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