Question 1013 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
Answer$\int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\sqrt{\text{x}}}+\frac{1}{\sqrt{\text{x}}\text{x}}\Big)\text{dx}$
$=\int\text{x}^{\frac{-1}{2}}+\int\text{x}^{\frac{-3}{2}}\text{dx}$
$=2\text{x}^{\frac{1}{2}}-2\text{x}^{\frac{-1}{2}}+\text{C}$
$=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
$\therefore\ \int\frac{1}{\sqrt{\text{x}}}\Big(1+\frac{1}{\text{x}}\Big)\text{dx}=2\sqrt{\text{x}}-\frac{2}{\sqrt{\text{x}}}+\text{C}$
View full question & answer→Question 1023 Marks
Evaluate the integral in Exercise:$\int^{2}_{0}\text{x}\sqrt{\text{x}+2}\ (\text{put}\ \text{x}+2=\text{t}^{2})$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{2}\text{x}\sqrt{\text{x}+2}\ \text{dx}$
$\text{putting}\sqrt{\text{x}+2}=\text{t}\ \Rightarrow\ \text{x}+2=\text{t}^{2}\ \Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{t}\ \Rightarrow\text{dx}=2\text{t}\ \text{dt}$
$\text{Limits of integration}\ \text{when}\ \text{x}=0,\text{t}=\sqrt{2}\ \text{and}\ \text{when}\text{x}=2,\text{t}=\sqrt{4}=2$
$\therefore\text{from}\ \text{eq}.\text{(i)},\ \text{I}=\int\limits_{\sqrt{2}}^{2}\text{t}^{2}(\text{t}^{2}-2)\text{t}.2\text{t}\ \text{dt}=2\int\limits_{\sqrt{2}}^{2}\text{t}^{2}(\text{t}^{2}-2)\text{dt}=2\int\limits_{\sqrt{2}}^{2}(\text{t}^{2}-2\text{t}^{2})\text{dt} $
$=2\Bigg[\bigg(\frac{\text{t}^{5}}{5}\bigg)^{2}_{\sqrt{2}}-2\bigg(\frac{\text{t}^{3}}{3}\bigg)^{2}_{\sqrt{2}}\Bigg]=2\bigg[\frac{1}{5}\Big(2^{5}-\big(\sqrt{2}\big)^{5}\Big)-\frac{2}{3}\Big(2^{3}-\big(\sqrt{2}\big)^{3}\Big)\bigg]$
$=2\bigg[\frac{1}{5}\big(32-4\sqrt{2}\big)-\frac{2}{3}\big(8-2\sqrt{2}\big)\bigg]=2\bigg[\frac{32}{5}-\frac{4\sqrt{2}}{5}-\frac{16}{3}+\frac{4\sqrt{2}}{3}\bigg]=2\bigg[\frac{96-12\sqrt{2}-80+20\sqrt{2}}{15}\bigg]$
$=\frac{2}{15}\big(16+8\sqrt{2}\big)=\frac{16\sqrt{2}}{15}\big(\sqrt{2}+1\big)$
View full question & answer→Question 1033 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{5}_{-5}|\text{x}+2|\ \text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{-5}^{5}|\text{x}+2|\ \text{dx}$
$\text{putting}\ \ \text{x}+2=0\ \ \Rightarrow=-2\in(-5,5)$
$\therefore\ \ \text{From eq. (i)},\ \text{I}=\int\limits_{-5}^{-2}|\text{x}+2|\ \text{dx}+\int\limits_{-2}^{5}|\text{x}+2|\ \text{dx}=\int\limits_{-5}^{-2}-\text{(x}+2 )\ \text{dx}+\int\limits_{-2}^{5}\text{(x}+2)\ \text{dx}$
$=-\bigg(\frac{\text{x}^{2}}{2}+2\text{x}\bigg)^{-2}_{-5}+\bigg(\frac{\text{x}^{2}}{2}+2\text{x}\bigg)^{5}_{-2}=-\bigg[\bigg(\frac{4}{2}-4\bigg)-\bigg(\frac{25}{2}-10\bigg)\bigg]+\bigg[\bigg(\frac{25}{2}-10\bigg)-\bigg(\frac{4}{2}-4\bigg)\bigg]$
$=-\bigg(-2-\frac{5}{2}\bigg)+\bigg(\frac{45}{2}+2\bigg)=2+\frac{5}{2}+\frac{45}{2}+2=4+25=29$
View full question & answer→Question 1043 Marks
Evaluate the following integrals:
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$
Answer$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2(\text{xe}^\text{x})}\text{ dx}$ Let $\text{xe}^\text{x}=\text{t}$ $\Rightarrow(1.\text{e}^\text{x}+\text{xe}^\text{x})=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{dx}=\text{dt}$ Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\sin^2({\text{xe}^\text{x}})}=\text{dx}$$=\int\frac{\text{dt}}{\sin^2\text{t}}$
$=\int\text{cosec}^2\text{t}\text{ dt}$
$=-\cot(\text{t})+\text{C}$
$=-\cot(\text{xe}^\text{x})+\text{C}$
View full question & answer→Question 1053 Marks
Evaluate the following integrals:
$\int\text{x e}^{\text{x}^2}=\text{dx}$
AnswerLet $\text{I}=\int\text{x e}^{\text{x}^2}=\text{dx}\ ....(1)$ Let $\text{x}^2=\text{t}$ then, $\text{d}(\text{x}^2)=\text{dt}$ $\Rightarrow2\text{x dx}=\text{dt}$ $\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$ Putting $\text{x}^2=\text{t}$ and $\text{x dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{e}^\text{t}\frac{\text{dt}}{2}$
$=\frac{1}{2}\text{e}^\text{t}+\text{C}$
$=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
$\text{I}=\frac{1}{2}\text{e}^{\text{x}^2}+\text{C}$
View full question & answer→Question 1063 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{2}\text{e}^{2\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{2\text{x}^2}\big)\text{dx}$
AnswerLet $\text{I}=\int_{1}^\limits{2}\text{e}^{2\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{2\text{x}^2}\big)\text{dx}$ Then,
$\text{I}=\int_{1}^\limits{2}\text{e}^{2\text{x}} \frac{1}{\text{x}}-\int_{1}^\limits{2}\text{e}^{2\text{x}} \frac{1}{2\text{x}^2}\text{ dx}$
Integrating first term by parts,
$\Rightarrow\text{I}=\bigg\{\Big[\frac{\text{e}^{2\text{x}}}{2\text{x}}\Big]^2_1-\int_{1}^\limits{2}-\text{e}^{2\text{x}} \frac{1}{2\text{x}^2}\bigg\}-\int_{1}^\limits{2}\text{e}^{2\text{x}} \frac{1}{2\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\Big[\frac{\text{e}^{2\text{x}}}{2\text{x}}\Big]^2_1$
$\Rightarrow\text{I}=\frac{\text{e}^4}{4}-\frac{\text{e}^2}{2}$
$\Rightarrow\text{I}=\frac{\text{e}^4-2\text{e}^2}{4}$
View full question & answer→Question 1073 Marks
Prove the following Exercise:
$\int^{\frac{\pi}{2}}_{0}\sin^{3}\text{x dx}=\frac{2}{3}$
Answer$\text{Let I}=\int^{\frac{\pi}{2}}_{0}\sin^{3}\text{x dx}$
$\text{I}=\int^{\frac{\pi}{2}}_{0}\sin^{2}\text{x.}\sin\text{x dx}$
$=\int^{\frac{\pi}{2}}_{0}\Big(1-\cos^{2}\text{x}\Big)\sin\text{x dx}$
$=\int^{\frac{\pi}{2}}_{0}\sin\text{x dx}-\int^{\frac{\pi}{2}}_{0}\cos^{2}\text{x}.\sin\text{x dx}$
$=\Big[-\cos\text{x}\Big]^{\frac{\pi}{0}}_{0}-\Big[\frac{\cos^{3}\text{x}}{3}\Big]^{\frac{\pi}{2}}_{0}$
$=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}$
Hence, the given result is Proved.
View full question & answer→Question 1083 Marks
Evaluate the following:
$\int\frac{\text{dx}}{\text{x}\sqrt{\text{x}^4}-1}$
Hint: Put $\text{x}^2=\sec\theta$
AnswerLet $\text{I}=\int\frac{\text{dx}}{\text{x}\sqrt{\text{x}^4}-1}$
Put $\text{x}^2=\sec\theta\Rightarrow\theta=\sec^{-1}\text{x}^2$
$\Rightarrow 2\text{xdx} =\sec\theta.\tan\theta \text{d}\theta$
$\therefore\ \text{I}=\frac{1}{2}\int\frac{\sec\theta\cdot\tan\theta}{\sec\theta\cdot\tan\theta}\text{d}\theta$ $=\frac{1}{2}\int\text{d}\theta=\frac{1}{2}\theta+\text{C}$
$=\frac{1}{2}\sec^{-1}(\text{x}^2)+\text{C}$
View full question & answer→Question 1093 Marks
Write a value of $\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}$
We know that,
$\int\text{e}^{\text{x}}\int\text{f}(\text{x})+\text{f}'(\text{x})=\text{e}^{\text{x}}\text{f}(\text{x})+\text{C}$
Hence, $\text{f}'(\text{x})=-\frac{1}{\text{x}^2}$
Then, $\int\text{e}^{\text{ax}}\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{dx}=\text{e}^{\text{x}}\cdot\frac{1}{\text{x}}+\text{C}$
$\therefore\ \text{I}=\text{e}^{\text{x}}\cdot\frac{1}{\text{x}}+\text{C}$
View full question & answer→Question 1103 Marks
Evaluate the following definite integrals:
$\int_{1}^\limits{\text{e}}\frac{\log\text{x}}{\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{1}^\limits{\text{e}}\frac{\log\text{x}}{\text{x}}\text{ dx}$
Let $\log\text{x}=\text{u}$
$\Rightarrow\frac{1}{\text{x}}=\text{dx}=\text{du}$
$\therefore\text{ I}=\int\text{u}\text{ du}$
$\Rightarrow\text{I}=\Big[\frac{\text{u}^2}{2}\Big]$
$\Rightarrow\text{I}=\Big[\frac{(\log\text{x})}{2}\Big]^{\text{e}}_1$
$\Rightarrow\text{I}=\frac{1}{2}-0$
$\Rightarrow\text{I}=\frac{1}{2}$
View full question & answer→Question 1113 Marks
Evaluate the following integrals:
$\int\limits^{\sqrt{2}}_0\big[\text{x}^2\big]\text{dx}$
AnswerWe have,
$\text{I}=\int\limits^{\sqrt{2}}_0\big[\text{x}^2\big]\text{dx}$
$=\int\limits^{1}_0\big[\text{x}^2\big]\text{dx}+\int\limits^{\sqrt{2}}_1\big[\text{x}^2\big]\text{dx}$
$=\int\limits^{1}_0(0)\text{dx}+\int\limits^{\sqrt{2}}_1(1)\text{dx}$ $\begin{pmatrix}\because\big[\text{x}\big]^2=\begin{cases}0,&0<\text{x}<1\\1,&1<\text{x}<\sqrt{2}\end{cases}\end{pmatrix}$
$=0+\big[\text{x}\big]^{\sqrt{2}}_1$
$=\sqrt{2}-1$
View full question & answer→Question 1123 Marks
Evaluate the following integrals:
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
AnswerWe have,
$\text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}$
Let $\text{f}(\text{x})=\text{x}\cos^2\text{x}$
$\Rightarrow \text{f}(-\text{x})= (-\text{x})\cos^2(-\text{x})$
$= - \text{x}\cos^2\text{x}$
$\therefore \text{f}(-\text{x})=-\text{f}(\text{x})$
i.e., f(x) is odd function.
We know that $\int\limits^\text{a}_{-\text{a}}\text{f}(\text{x})\text{dx} = 0, $ if f(x) is odd function.
$\therefore\ \text{I}=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\text{x}\cos^2\text{x}\text{ dx}=0$
View full question & answer→Question 1133 Marks
Integrate the following integrals:
$\int\sin4\text{x}\cos7\text{x dx}$
Answer$\int\sin4\text{x}\cos7\text{x dx}$
$=\frac{1}{2}\int2\cos7\text{x}\sin4\text{x dx}$
$=\frac{1}{2}\int\big[\sin(7\text{x}+4\text{x})-\sin(7\text{x}-4\text{x})\big]\text{dx}$ $[\therefore2\cos\text{A}\sin\text{B}=\sin(\text{A}+\text{B})-\sin(\text{A}-\text{B})\big]$
$=\frac{1}{2}\int\big(\sin(11\text{x})-\sin(3\text{x})\big)\text{dx}$
$=\frac{1}{2}\Big[-\frac{\cos(11\text{x})}{11}+\frac{\cos(3\text{x})}{3}\Big]+\text{c}$
$=-\frac{\cos(11\text{x})}{22}+\frac{\cos(3\text{x})}{6}$
View full question & answer→Question 1143 Marks
Evaluate the following:
$\int\frac{\sqrt{\text{x}}}{\sqrt{\text{a}^3-\text{x}^3}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{\text{x}}}{\sqrt{\text{a}^3-\text{x}^3}}\text{dx}$ $=\int\frac{\sqrt{\text{x}}}{\sqrt{\Big(\text{a}^{\frac{3}{2}}\Big)^2-\Big(\text{x}^{\frac{3}{2}}\Big)^2}}$
Put $=\text{x}^{\frac{3}{2}}=\text{t}\Rightarrow\frac{3}{2}\text{x}^{\frac{1}{2}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{2}{3}\int\frac{\text{dt}}{\sqrt{\Big(\text{a}^{\frac{3}{2}}\Big)^2}-\text{t}^2}$ $=\frac{2}{3}\sin^{-1}\frac{\text{t}}{\text{a}^{\frac{3}{2}}}+\text{C}$
$=\frac{2}{3}\sin^{-1}\frac{\text{x}^{\frac{3}{2}}}{\text{a}^{\frac{3}{2}}}+\text{C}$ $=\frac{2}{3}\sin^{-1}\sqrt{\frac{\text{x}^3}{\text{a}^3}}+\text{C}$
View full question & answer→Question 1153 Marks
Write a value of $\int\frac{1}{\text{x}(\log\text{x})^{\text{n}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}(\log\text{x})^{\text{n}}}\text{ dx}$
Let $\log\text{x}=\text{t}$
$\frac{1}{\text{x}}\text{ dx}=\text{dt}$
$\text{dx}=\text{xdt}$
$\therefore\ \text{I}=\int\frac{1}{\text{t}^{\text{n}}}\text{ dt}$
$=\frac{\text{t}^{-\text{n}+1}}{-\text{n}+1}+\text{C}$
$\text{I}=\frac{(\log\text{x})^{1-\text{n}}}{1-\text{n}}+\text{C}$
View full question & answer→Question 1163 Marks
Evaluate the following:
$\int\limits^1_0\frac{\text{dx}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\limits^1_0\frac{\text{dx}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$
$=\int\limits^1_0\frac{\text{dx}}{\text{e}^\text{x}+\frac{1}{\text{e}^\text{x}}}$
$=\int\limits^0_1\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{dx}$
Put $\text{e}^\text{x}=\text{t}$
$\text{e}^\text{x}\text{dx}=\text{dt}$
Substituting ex = t and ex dx = dt
$\therefore\ \text{I}=\int\limits^\text{e}_1\frac{\text{dt}}{1+\text{t}^2}$
$=\big[\tan^{-1}\text{t}\big]^\text{e}_\text{1}$ $\Big[\because\int\frac{1}{1+\text{x}^2}\text{dx}=\tan^{-1}\text{x}+\text{C}\Big]$
$=\tan^{-1}\text{e}-\tan^{-1}1$
$=\tan^{-1}\text{e}-\frac{\pi}{4}$ $\Big[\because\tan^{-1}1=\frac{\pi}{4}\Big]$
View full question & answer→Question 1173 Marks
Evaluate the following:
$\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$ $=\int\sqrt{-(\text{x}^2-2\text{ax})\text{dx}}$
$=\int\sqrt{-(\text{x}^2-2\text{ax}+\text{a}^2-\text{a}^2)}\text{dx}$ $=\int\sqrt{-(\text{x}-\text{a})^2-\text{a}^2}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}-\text{a})^2}\text{dx}$
$=\frac{\text{x}-\text{a}}{2}\sqrt{\text{a}^2-(\text{x}-\text{a})^2}+\frac{\text{a}^2}{2}\sin^{-1}\Big(\frac{\text{x}-\text{a}}{\text{x}}\Big)+\text{C}$
$=\frac{\text{x}-\text{a}}{2}\sqrt{2\text{ax}-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\Big(\frac{\text{x}-\text{a}}{\text{a}}\Big)+\text{C}$
View full question & answer→Question 1183 Marks
Evaluate the following integrals:
$\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$
Answer$\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$ Let $\text{x}+\log\text{x}=\text{t}$ $\Rightarrow\Big(1+\frac{1}{\text{x}}\Big)=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{dx}=\text{dt}$Now, $\int\Big(\frac{\text{x}+1}{\text{x}}\Big)(\text{x}+\log\text{x})^2\text{dx}$
$=\int\text{t}^2\text{dt}$
$=\frac{\text{t}^2}{3}+\text{C}$
$=\frac{(\text{x}+\log\text{x})^3}{3}+\text{C}$
View full question & answer→Question 1193 Marks
Evaluate the following:
$\int\sqrt{1+\sin\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{1+\sin\text{x}}\text{dx}$
$=\int\sqrt{\sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}+2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}\text{dx}$ $\Big[\because\ \sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}=1\Big]$
$=\int\sqrt{\big(\sin\frac{\text{x}}{2}+\cos\frac{\text{x}}{2}\big)^2}\text{dx}$
$=\int\Big(\sin\frac{\text{x}}{2}+\cos\frac{\text{x}}{2}\Big)\text{dx}$
$=-2\cos\frac{\text{x}}{2}+2\sin\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 1203 Marks
Evalute the following integrals:
$\int\frac{1}{\sqrt{1-\cos2\text{x}}}\text{dx}$
Answer$\int\frac{1}{\sqrt{1-\cos2\text{x}}}\text{dx}$$\int\frac{1}{\sqrt{2\sin^2\text{x}}}\text{dx}\ \big[\because 1-\cos 2\text{x}=2\sin^2\text{x}\big]$
$=\frac{1}{\sqrt{2}}\int\text{cosec x dx}$
$=\frac{1}{\sqrt{2}}\text{ln}|\text{cosec x}-\cot\text{x}|=\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\frac{1}{\sin\text{x}}-\frac{\cos\text{x}}{\sin\text{x}}\Big|+\text{C}$
$=\frac{1}{\sqrt{2}}\text{ln}\bigg|\frac{2\sin^2\frac{\text{x}}{2}}{\sin\text{x}}\bigg|+\text{C} \Big[\because 1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Bigg|\frac{2\sin^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}\Bigg|+\text{C}\ \Big[\because\sin\text{x}=2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}\Big]$
$=\frac{1}{\sqrt{2}}\text{ln}\Big|\tan\frac{\text{x}}{2}\Big|+\text{C}$
View full question & answer→Question 1213 Marks
Evaluate the following integrals:
$\int\frac{1}{1-\sin\text{x}}\text{dx}$
Answer$\int\frac{1}{1-\sin\text{x}}\text{dx}$$=\int\frac{(1+\sin\text{x})}{(1-\sin\text{x})\times(1+\sin\text{x})}\text{dx}$
$=\Big(\frac{1+\sin\text{x}}{1-\sin^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1+\sin\text{x}}{\cos^2\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\Big)\text{dx}$
$=\int(\sec^2\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
$=\tan\text{x}+\sec\text{x}+\text{C}$
View full question & answer→Question 1223 Marks
Write a value of $\int\frac{\sin\text{x}}{\cos^3\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}}{\cos^3\text{x}}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x dx}=\text{dt}$
$\sin\text{x dx}=-\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{\text{t}^3}$
$=-\int\text{t}^{-3}\text{dt}$
$=-\Big[\frac{\text{t}^{-3+1}}{-3+1}\Big]+\text{C}$
$=\frac{1}{2\text{t}^2}+\text{C}$
$=\frac{1}{2\cos^2\text{x}}+\text{C}$ $(\because\text{t}=\cos\text{x})$
$=\frac{1}{2}\sec^2\text{x}+\text{C}$
View full question & answer→Question 1233 Marks
Evaluate the following integrals:$\int\frac{1}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-4}}\text{ dx}$
Answer$\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\Big(\text{x}^{\frac{1}{3}}\Big)^2-2^2}}$
Let $\text{x}^{\frac{1}{3}}=\text{t}$
$\Rightarrow\frac{1}{3}\text{x}^{\frac{-2}{3}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{1}{3\text{x}^{\frac{2}{3}}}\text{ dx}=\text{dt}$
$\Rightarrow\frac{\text{dx}}{\text{x}^{\frac{2}{3}}}=3\text{dt}$
Now, $\int\frac{\text{dx}}{\text{x}^{\frac{2}{3}}\sqrt{\text{x}^{\frac{2}{3}}-2^2}}$
$=3\int\frac{\text{dt}}{\sqrt{\text{t}^2-2^2}}$
$=3\log\Big|\text{t}+\sqrt{\text{t}^2-2^2}\Big|+\text{C}$
$=3\log\Bigg|\text{x}^{\frac{1}{3}}+\sqrt{\text{x}^{\frac{2}{3}}-4}\Bigg|+\text{C}$
View full question & answer→Question 1243 Marks
Evaluate the following:
$\int\frac{\text{dx}}{1+\cos\text{x}}$
AnswerLet $\int\frac{\text{dx}}{1+\cos\text{x}}$
$=\int\frac{\text{x}}{2\cos^2\frac{\text{x}}{2}}$ $\Big[\because\ 1+\cos\text{A}=2\cos^2\frac{\text{A}}{2}\Big]$
$=\frac{1}{2}\int\frac{1}{\cos^2\frac{\text{x}}{2}}\text{dx}$
$=\frac{1}{2}\int\sec^2\frac{\text{x}}{2}\text{dx}$
$=\frac{1}{2}\cdot\tan\frac{\text{x}}{2}\cdot2+\text{C}$ $\big[\int\sec^2\text{x dx}=\tan\text{x}\big]$
$=\tan\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 1253 Marks
Evalute the following integrals:
$\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\text{dx}$
AnswerLet $\int\frac{\sec^2\text{x}}{\tan\text{x}+2}\text{dx}\ .....\text{(i)}$
Let $\tan\text{x}+2=\text{t}$ then,
$\text{d}(\tan\text{x}+2)=\text{dt}$
$\Rightarrow\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{1}{\sec^2\text{x}}\text{dt}$
Putting $\tan\text{x}+2=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\sec^2\text{x}}$ In equation (i), we get,
$\text{I}=\int\frac{\sec^2\text{x}}{\text{t}}\times\frac{1}{\sec^2\text{x}}\text{dt}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\tan\text{x}+2|+\text{C}$
$\Rightarrow\text{I}=\log|\tan\text{x}+2|+\text{C}$
View full question & answer→Question 1263 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}(\tan\text{x}-\log\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\tan\text{x}-\log\cos\text{x})\text{dx}$$=\int\text{e}^{\text{x}}\tan\text{x dx}-\int\text{e}^{\text{x}}\log\cos\text{x dx}$
Integrating by parts
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\Big\{\text{e}^{\text{x}}\log\cos\text{x}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}\log\cos\text{x}\Big)\text{dx}\Big\}$
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\big\{\text{e}^{\text{x}}\log\cos\text{x}+\int\text{e}^{\text{x}}\tan\text{x dx}\big\}$
$=\int\text{e}^{\text{x}}\tan\text{x dx}-\text{e}^{\text{x}}\log\cos\text{x}-\int\text{e}^{\text{x}}\tan\text{x dx}+\text{C}$
$=-\text{e}^{\text{x}}\log\cos\text{x}+\text{C}$
$=\text{e}^{\text{x}}\log\sec\text{x}+\text{C}$ $\big[\because\log\sec\text{x}=-\log\cos\text{x}\big]$
View full question & answer→Question 1273 Marks
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{\frac{\pi}{2}}_{0}\frac{\cos^{5}\text{x}\ \text{dx}}{\sin^{5}\text{x}+\cos^{5}\text{x}}$
Answer$\text{Let}\text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos^{5}\text{x}\ }{\sin^{5}\text{x}+\cos^{5}\text{x}}\text{dx}$ $\Rightarrow\ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos^{5}\bigg(\frac{\pi}{2}-\text{x}\bigg)}{\sin^{5}\bigg(\frac{\pi}{2}-\text{x}\bigg)+\cos^{5}\bigg(\frac{\pi}{2}-\text{x}\bigg)}\text{dx}\ \ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\ \text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{(a}-\text{x)}\text{dx}=\bigg]$ $\Rightarrow\ \ \text{I}=\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^{5}\text{x}}{\cos^{5}\text{x}+\sin^{5}\text{x}}\text{dx}$Adding eq. (i) and (ii),
$21=\int\limits_{0}^{\frac{\pi}{2}}\bigg(\frac{\cos^{5}\text{x}}{\sin^{5}\text{x}+\cos^{5}\text{x}}+\frac{\sin^{5}\text{x}}{\cos^{5}\text{x}+\sin^{5}\text{x}}\bigg)\text{dx}=\int\limits_{0}^{\frac{\pi}{2}}\bigg(\frac{\cos^{5}\text{x}+\sin^{5}\text{x}}{\sin^{5}\text{x}+\cos{5}\text{x}}\bigg)\text{dx}$ $\Rightarrow\ \ 21=\int\limits_{0}^{\frac{\pi}{2}}1\ \text{dx}=\bigg(\text{x}^{\frac{\pi}{2}}_{0}\bigg)\ \Rightarrow21=\frac{\pi}{2}\ \Rightarrow\text{I}=\frac{\pi}{4}$
View full question & answer→Question 1283 Marks
Find the integrals of the functions in Exercises:
$\frac{\cos\text{x}}{1+\cos\text{x}}$
Answer$\frac{\cos\text{x}}{1+\cos\text{x}}=\frac{\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}$ $\ \ \ \ \ \ \ \ \bigg[\cos\text{x}=\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}\text{ and }\cos\text{x}=2\cos^2\frac{\text{x}}{2}-1\bigg]$
$=\frac{1}{2}\bigg[1-\tan^2\frac{\text{x}}{2}\bigg]$
$\therefore\int\frac{\cos\text{x}}{1+\cos\text{x}}\text{ dx}=\frac{1}{2}\int\bigg(1-\tan^2\frac{\text{x}}{2}\bigg)\text{ dx}$
$=\frac{1}{2}\int\bigg(1-\sec^2\frac{\text{x}}{2}+1\bigg)\text{dx}$
$=\frac{1}{2}\int\bigg(2-\sec^2\frac{\text{x}}{2}\bigg)\text{dx}$
$=\frac{1}{2}\Bigg[2\text{x}-\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\Bigg]+\text{C}$
$=\text{x}-\tan\frac{\text{x}}{2}+\text{C}$
View full question & answer→Question 1293 Marks
Evaluate the following integrals:
$\int\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\text{dx}$
Answer$\int\Big(\frac{\cos^2\text{x}-\sin^2\text{x}}{\sqrt{1+\cos4\text{x}}}\Big)\text{x}$
$=\int\frac{\cos(2\text{x})}{\sqrt{2\cos^2(2\text{x})}}\text{dx}$ $\Big[\therefore1+\cos\text{A}=2\cos^2\Big(\frac{\text{A}}{2}\Big)\text{ and }\cos^2\text{A}-\sin^2\text{A}=\cos\text{2A}\Big]$
$=\frac{1}{\sqrt2}\int\Big(\frac{\cos2\text{x}}{\cos\text{2x}}\Big)\text{dx}$
$=\frac{1}{\sqrt{2}}[\text{x}]+\text{C}$
$=\frac{\text{x}}{\sqrt{2}}+\text{C}$
View full question & answer→Question 1303 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx}$
Here, $\text{f(x)}=\log\sin\text{x}$ Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{f}'\text{(x)}=\cot\text{x}$
let $\text{e}^{\text{x}}\log\sin\text{x = t}$
Diff. both sides w.r.t x
$\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\times\frac{1}{\sin\text{x}}\times\cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big[\text{e}^{\text{x}}\log(\sin\text{x})+\text{e}^{\text{x}}\cot\text{x}\big]\text{dx = dt}$
$\Rightarrow\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx = dt}$
$\therefore \int\text{e}^{\text{x}}(\cot\text{x}+\log\sin\text{x})\text{dx} =\int\text{dt}$
$=\text{t + C}$
$=\text{e}^{\text{x}}\log\sin\text{x}+\text{C}$
View full question & answer→Question 1313 Marks
Evaluate the following integrals:
$\int\tan^32\text{x}\sec2\text{x dx}$
Answer$\int\tan^32\text{x}\sec2\text{x}=\tan^22\text{x}\tan2\text{x}\sec2\text{x}$
$=\big(\sec^22\text{x}-1\big)\tan2\text{x}\sec2\text{x}$
$=\sec^22\text{x}\tan2\text{x}\sec2\text{x}-\tan2\text{x}\sec2\text{x}$
$\therefore\ \int\tan^32\text{x}\sec2\text{x}\text{ dx}$
$=\int\sec^22\text{x}\tan2\text{x}\sec2\text{x dx}-\int\tan2\text{x}\sec2\text{x dx}$
$=\int\sec^22\text{x}\tan2\text{x}\sec2\text{x dx}-\frac{\sec2\text{x}}{2}+\text{C}$
Let $2\text{x}=\text{t}$
$\therefore\ 2\sec2\text{x}\tan2\text{x dx}=\text{dt}$
$\therefore\ \int\tan^32\text{x}\sec2\text{x dx}=\frac{1}{2}\int\text{t}^2\text{dt}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{\text{t}^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
$=\frac{(\sec2\text{x})^3}{6}-\frac{\sec2\text{x}}{2}+\text{C}$
View full question & answer→Question 1323 Marks
Evaluate the following integrals:
$\int_{4}^\limits{12}\text{x}(\text{x}-4)^{\frac{1}{3}}\text{dx}$
AnswerLet $\text{I}=\int_{4}^\limits{12}\text{x}(\text{x}-4)^{\frac{1}{3}}\text{dx}$
Let $\text{x}-4=\text{t}$ Then, $\text{dx}=\text{dt}$
When $\text{x}=4,\text{t}=0$ and $\text{x}=12,\text{t}=8$
$\therefore\ \text{I}=\int\limits^8_0(\text{t}+4)\text{t}^{\frac{1}{3}}\text{dt}$
$\Rightarrow\text{I}=\int\limits^8_0\Big(\text{t}^{\frac{4}{3}}+4\text{t}^{\frac{1}{3}}\Big)\text{dt}$
$\Rightarrow\text{I}=\Big[\frac{3}{7}\text{t}^{\frac{7}{3}}+\frac{3}{1}\text{t}^{\frac{4}{3}}\Big]^8_0$
$\Rightarrow\text{I}=\frac{384}{7}+48$
$\Rightarrow\text{I}=\frac{720}{7}$
View full question & answer→Question 1333 Marks
Evaluate the following integrals:
$\int\limits^2_0\text{x}\sqrt{2-\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\limits^2_0\text{x}\sqrt{2-\text{x}}\text{ dx}$
$=\int\limits^2_0(2-\text{x})\sqrt{2-2+\text{x}}\text{ dx}$
$=\int\limits^2_0(2-\text{x})\sqrt{\text{x}}\text{ dx}$
$=\int\limits^2_0\big(2\sqrt{\text{x}}-\text{x}\sqrt{\text{x}}\big)\text{dx}$
$=\int\limits^2_0\Big(2\text{x}^{\frac{1}{2}}-\text{x}^{\frac{3}{2}}\Big)\text{dx}$
$=\Bigg[2\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\text{x}^{\frac{5}{2}}}{\frac{5}{2}}\Bigg]$
$=\bigg[\frac{4}{3}\text{x}^{\frac{3}{2}}-\frac{2}{5}\text{x}^{\frac{5}{2}}\bigg]^2_0$
$=\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}$
$=\frac{16\sqrt{2}}{15}$
View full question & answer→Question 1343 Marks
Evaluate the following integrals:
$\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$Let $\tan^{-1}\text{x}=\text{t}$ Then, $\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
When $\text{x}=0,\text{t}=0$ and $\text{x}=1,\text{t}=\frac{\pi}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{1}\frac{\sqrt{\tan^{-1}\text{x}}}{1+\text{x}^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\sqrt{\text{t}}\text{ dt}$
$\Rightarrow\text{I}=\Bigg[\frac{2\text{t}^{\frac{3}{2}}}{3}\Bigg]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{2}{3}\Big(\frac{\pi}{4}\Big)^{\frac{3}{2}}$
$\Rightarrow\text{I}=\frac{1}{12}\pi^{\frac{3}{2}}$
View full question & answer→Question 1353 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}}\Big(\text{x}^3-\frac{2}{\text{x}}\Big)\text{dx}$
Answer$\int\sqrt{\text{x}}\Big(\text{x}^3-\frac{2}{\text{x}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{7}{2}}-\frac{2}{\sqrt{\text{x}}}\Big)\text{dx}$
$=\int\Big(\text{x}^{\frac{7}{2}}-{2}{\text{x}^{-\frac{1}{2}}}\Big)\text{dx}$
$=\frac{\text{x}^{\frac{7}{2}+1}}{\frac{7}{2}+1}-2\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=\frac{2}{9}\text{x}^{\frac{9}{2}}-4\text{x}^{\frac{1}{2}}+\text{C}$
$=\frac{2}{9}\text{x}^{\frac{9}{2}}-4\sqrt{\text{x}}+\text{C}$
View full question & answer→Question 1363 Marks
Evaluate the following integrals:$\int\frac{\sin8\text{x}}{\sqrt{9+\sin^44\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin8\text{x}}{\sqrt{9+(\sin4\text{x})^4}}\text{ dx}$
Let $\sin^24\text{x}=\text{t}$
$\Rightarrow2\sin4\text{x}.\cos4\text{x}(4)\text{dx}=\text{dt}$
$\Rightarrow4\sin8\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin8\text{x}\text{ dx}=\frac{\text{dt}}{4}$
$\text{I}=\frac{1}{4}\int\frac{\text{dt}}{\sqrt{(3)^2+\text{t}^2}}$
$\text{I}=\frac{1}{4}\log\Big|\text{t}+\sqrt{(3)^2+\text{t}^2}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}\Big]$
$\text{I}=\frac{1}{4}\log\Big|\sin^24\text{x}+\sqrt{9+\sin^44\text{x}}\Big|+\text{C}$
View full question & answer→Question 1373 Marks
Evalute the following integrals:
$\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
AnswerNote: Here, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.
Let $\text{I}=\int\frac{1}{\text{x}\log\text{x}\log(\log\text{x})}\text{dx}$
Putting $\log(\log\text{x})=\text{t}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{x}\log\text{x}}\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|\log(\log\text{x})|+\text{C}\ \big[\because\text{t}=\log(\log\text{x})\big]$
View full question & answer→Question 1383 Marks
Evaluate the following integrals:
$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
Answer$\int\frac{\sin2\text{x}}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$\text{Let a}+\text{b}\cos2\text{x}=\text{t}$
$\Rightarrow-\text{b}\sin(2\text{x})\text{dx}\times2=\text{dt}$
$\Rightarrow\sin(2\text{x})\text{dx}=\frac{-\text{dt}}{2\text{b}}$
$\text{Now,}\int\frac{\sin(2\text{x})}{(\text{a}+\text{b}\cos2\text{x})^2}\text{dx}$
$=-\frac{1}{2\text{b}}\int\frac{\text{dt}}{\text{t}^2}$
$=\frac{-1}{2\text{b}}\int\text{t}^{-2}\text{dt}$
$=\frac{-1}{2\text{b}}\Big[\frac{\text{t}^{-2+1}}{-2+1}\Big]+\text{C}$
$=\frac{1}{2\text{b}}\times\frac{1}{\text{t}}+\text{C}$
$=\frac{1}{2\text{b}(\text{a}+\text{b}\cos2\text{x})}+\text{C}$
View full question & answer→Question 1393 Marks
Evaluate the following integrals:
$\int\limits^{2\pi}_0\cos^{-1}(\cos\text{x})\text{dx}$
Answer$\int\limits^{2\pi}_0\cos^{-1}(\cos\text{x})\text{dx}$
$=\int\limits^{\pi}_0\cos^{-1}(\cos\text{x})\text{dx}+\int\limits^{2\pi}_\pi\cos^{-1}(\cos\text{x})\text{dx}$
$=\int\limits^{\pi}_0\text{x}\text{ dx}+\int\limits^{2\pi}_\pi(2\pi-\text{x})\text{dx}$ $\big[{\pi}\leq\text{x}\leq2\pi\Rightarrow-2\pi\leq-\text{x}\leq-{\pi}\Rightarrow0\leq2\pi-\text{x}\leq{\pi}\big]$
$=\Big[\frac{\text{x}^2}{2}\Big]+\bigg[\frac{(2\pi-\text{x})}{2\times(-1)}\bigg]^{2\pi}_\pi$
$=\frac{1}{2}(\pi^2-0)-\frac{1}{2}(0-\pi^2)$
$=\frac{\pi^2}{2}+\frac{\pi^2}{2}$
$=\pi^2$
View full question & answer→Question 1403 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{1}\text{x}(1-\text{x})^5\text{ dx}$
AnswerWe have,
$\int_{0}^\limits{1}\text{x}(1-\text{x})^5\text{ dx}$
Expanding $(1 - x)^5$ by Binomial theorem,
$\therefore\ (1-\text{x})^5=1^5+{^5\text{C}_1}(-\text{x})+{^5\text{C}_2}(-\text{x})^2\\+{^5\text{C}_3}(-\text{x})^3+{^5\text{C}_4}(-\text{x})^4+{^5\text{C}_5}(-\text{x})^5$
$=1-5\text{x}+10\text{x}^2-10\text{x}^3+5\text{x}^4-\text{x}^5$
$=\int_{0}^\limits{1}\text{x}(1-5\text{x}+10\text{x}^2-10\text{x}^3+5\text{x}^4-\text{x}^5)\text{dx}$
$=\Big[\frac{\text{x}^2}{2}-\frac{5\text{x}^3}{3}+\frac{10\text{x}^4}{4}-\frac{10\text{x}^5}{5}+\frac{5\text{x}^6}{6}-\frac{\text{x}^7}{7}\Big]^1_0$
$=\frac{1}{2}-\frac{5}{3}+\frac{10}{4}-\frac{10}{5}+\frac{5}{6}-\frac{1}{7}$
$=\frac{1}{42}$
View full question & answer→Question 1413 Marks
Evaluate the definite integral in Exercise:
$\int_{0}^{1}(\text{x}\text{e}^{\text{x}}+\sin\frac{\pi\text{x}}{4})\text{dx}$
Answer$\text{Let}\ \text{I}=\int\limits_{0}^{1}\bigg(\text{xe}^{\text{x}}+\sin\frac{\pi\text{x}}{4}\bigg)\text{dx}$
$\int\bigg(\text{xe}^{\text{x}}+\sin\frac{\pi\text{x}}{4}\bigg)\text{dx}=\text{x}\int\text{e}^\text{x}\text{dx}-\int\left\{\bigg(\frac{\text{d}}{\text{dx}}\text{x}\bigg)\int\text{e}^\text{x}\text{dx}\right\}\text{dx}+\left\{\frac{-\cos\frac{\pi\text{x}}{4}}{\frac{\pi}{4}}\right\}$
$=\text{xe}^\text{x}-\int\text{e}^\text{x}\text{dx}-\frac{4}{\pi}\cos\frac{\pi\text{x}}{4}$
$=\text{xe}^\text{x}-\text{e}^\text{x}-\frac{4}{\pi}\cos\frac{\pi\text{x}}{4}$
$=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain
$\text{I}=\text{F}(1)-\text{F}(0)$
$=\bigg(1.\text{e}^{1}-\text{e}^{1}-\frac{4}{\pi}\cos\frac{\pi}{4}\bigg)-\bigg(0.\text{e}^{0}-\text{e}^{0}-\frac{4}{\pi}\cos0\bigg)$
$=\text{e}-\text{e}-\frac{4}{\pi}\bigg(\frac{1}{\sqrt{2}}\bigg)+1+\frac{4}{\pi}$
$=1+\frac{4}{\pi}-\frac{2\sqrt{2}}{\pi}$
View full question & answer→Question 1423 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\big[\sec\text{x}+\log(\sec\text{x}+\tan\text{x})\big]\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\big[\sec\text{x}+\log(\sec\text{x}+\tan\text{x})\big]\text{dx}$
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\int\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})\text{dx}$
Integrating by parts
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\text{e}^{\text{x}}\log(\sec\text{x}\tan\text{x})-\int\text{e}^{\text{dx}}\Big\{\frac{\text{d}}{\text{dx}}\log(\sec\text{x}+\tan\text{x})\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\sec\text{x dx}+\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})-\int\text{e}^{\text{x}}\sec\text{x dx}$
$=\text{e}^{\text{x}}\log(\sec\text{x}+\tan\text{x})+\text{C}$
View full question & answer→Question 1433 Marks
Evaluate the following integrals:
$\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int\text{x}\frac{\tan^{-1}\text{x}^2}{1+\text{x}^4}\text{ dx}\ ....(1)$ Let $\tan^{-1}\text{x}^2=\text{t}$ then, $\text{d}\big(\tan^{-1}\text{x}^2\big)=\text{dt}$ $\Rightarrow\frac{1\times2\text{x}}{1+(\text{x}^2)^2}\text{ dx}=\text{dt}$ $\Rightarrow\frac{1\times\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ Putting, $\tan^{-1}\text{x}^2=\text{t}$ and $\frac{\text{x}}{1+\text{x}^4}\text{ dx}=\frac{\text{dt}}{2}$ in equation (1),we get,
$\text{I}=\int\text{t}\frac{\text{dx}}{2}$
$=\frac{1}{2}\int\text{t dt}$
$=\frac{1}{2}\times\frac{\text{t}^2}{2}+\text{C}$
$\text{I}=\frac{\text{t}^2}{4}+\text{C}$
$=\frac{(\tan^{-1}\text{x}^2)^2}{4}+\text{C}$
$\text{I}=\frac{1}{4}\big(\tan^{-1}\text{x}^2\big)^2+\text{C}$
View full question & answer→Question 1443 Marks
Evaluate the following:
$\int\sqrt{5-2\text{x}+\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{5-2\text{x}+\text{x}^2}\text{dx}$
$=\int\sqrt{\text{x}^2-2\text{x}+1+4}\text{dx}$
$=\int\sqrt{(\text{x}-1)^2+(2)^2}\text{dx}$
Using $\int\sqrt{\text{x}^2+\text{a}^2}\text{dx}$ $=\frac{1}{2}\text{x}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\log\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|+\text{C},$ we get
$\int\sqrt{(\text{x}-1)^2+(2)^2}\text{dx}$ $=\frac{\text{x}-1}{2}\sqrt{2^2+(\text{x}+1)^2}+2\log\Big|\text{x}-1+\sqrt{2^2+(\text{x}-1)^2}\Big|+\text{C}$
$\Rightarrow\ \int\sqrt{5-2\text{x}+\text{x}^2}\text{dx}$ $=\frac{\text{x}-1}{2}\sqrt{5-2\text{x}+\text{x}^2}+2\log\Big|\text{x}-1+\sqrt{5-2\text{x}+\text{x}^2}\Big|+\text{C}$
View full question & answer→Question 1453 Marks
Evaluate the following integrals:
$\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
AnswerLet $\text{I}=\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)(1+\text{xe}^\text{x}-\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{\int(\text{x}+1)(1+\text{xe}^\text{x})}{\text(1+\text{xe}^\text{x})}\ \text{dx}-\int\frac{(\text{x}+1)(\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)}{\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\int\frac{(\text{x}+1)\text{e}^\text{x}}{\text{xe}^\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\log|\text{xe}^\text{x}|-\log|1+\text{xe}^\text{x}|+\text{C}$
$\therefore\text{I}=\log\Big|\frac{\text{xe}^\text{x}}{1+\text{xe}^\text{x}}\Big|+\text{C}$
View full question & answer→Question 1463 Marks
Evaluate the following integrals:
$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
Answer$\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$ Let $\text{x}+\tan^{-1}\text{x}=\text{t}$ $\Big(1+\frac{1}{1+\text{x}^2}\Big)=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\Big(\frac{\text{x}^2-1+1}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$ $\Rightarrow\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}=\text{dt}$Now, $\int5^{\text{x}+\tan^{-1}\text{x}}.\Big(\frac{\text{x}^2+2}{\text{x}^2+1}\Big)\text{dx}$
$=\int5^\text{t}\text{dt}$
$=\frac{5^\text{t}}{\log5}+\text{C}$
$=\frac{5^{\text{x}+\tan^{-1}\text{x}}}{\log5}+\text{C}$
View full question & answer→Question 1473 Marks
Write a value of $\int\log_\text{e}\text{x}\text{ dx}$
Answer$\int\log_\text{e}\text{x}\text{ dx}$
$=\int1\cdot\log_\text{e}\text{x dx}$
$=\log_\text{e}\text{x}\int1\text{ dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\log_\text{e}\text{x})\int1\text{dx}\Big\}\text{dx}$
$=\log_\text{e}\text{x}\int1\cdot\text{dx}-\int\frac{1}{\text{x}}\cdot\text{x dx}$
$=\log_\text{e}\text{x}\cdot\text{x}-\int\text{dx}$
$=\text{x}\log_\text{e}\text{x}-\text{x}+\text{C}$
$=\text{x}(\log_\text{e}\text{x}-1)+\text{C}$
View full question & answer→Question 1483 Marks
Evaluate the following integrals:
$\int (3\text{x}\sqrt{\text{x}}+4\sqrt{\text{x}}+5)\text{dx}$
Answer$\int(3\text{x}\sqrt{5}+4\sqrt{\text{x}}+5)\text{dx}$
$=\int3\text{x}\sqrt{5}\text{dx}+\int4\sqrt{\text{x}}\text{dx}+\int5\text{dx}$
$=\int3\text{x}^{\frac{3}{2}}\text{dx}+4\int\text{x}^{\frac{1}{2}}\text{dx}+5\int\text{dx}$
$=\frac{\text{x}\frac{3}{2}+1}{\frac{3}{2}+1}+\frac{4\text{x}^{\frac{1}{2}}}{\frac{1}{2}+1}+5\text{x}+\text{C}$
$=\frac{6}{5}\text{x}^{\frac{5}{2}}+\frac{8}{3}\text{x}^{\frac{3}{2}}+5\text{x}+\text{C}$
View full question & answer→Question 1493 Marks
Evaluate the following integrals:$\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{}x}\Big(\frac{1+\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\frac{1}{1+\cos\text{x}}+\frac{\sin\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\bigg(\frac{1}{2\cos\frac{\text{x}}{2}}+\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\text{e}^{\text{}x}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}$
Putting $\text{e}^{\text{x}}\tan\frac{\text{x}}{2}=\text{t}$
Diff. both sides w.r.t.x
$\text{e}^{\text{x}}.\tan\big(\frac{\text{x}}{2}\big)+\text{e}^{\text{x}}\times\frac{1}{2}\sec^{2}\frac{\text{x}}{2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}\big[\tan\frac{\text{x}}{2}+\frac{1}{2}\sec^2\big(\frac{\text{x}}{2}\big)\big]\text{dx}=\text{dt}$
$\therefore\int\text{e}^{\text{x}}\big(\frac{1}{2}\sec^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}\big)\text{dx}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\tan\big(\frac{\text{x}}{2}\big)+\text{C}$
View full question & answer→Question 1503 Marks
Evaluate the following definite integrals:
$\int_{0}^\limits{\pi}\frac{1}{1+\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int_{0}^\limits{\pi}\frac{1}{1+\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\pi}\frac{1-\sin\text{x}}{(1+\sin\text{x})(1-\sin\text{x})}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1-\sin\text{x}}{1+\sin^2\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\frac{1-\sin\text{x}}{\cos^2\text{x}}\text{ dx}$ $\big[\because\sin^2\text{x}+\cos^2\text{x}=1\big]$
$\Rightarrow\text{I}=\int_{0}^\limits{\pi}\sec^2\text{x}-\sec\text{x}\tan\text{x}\text{ dx}$
$\Rightarrow\text{I}=\big[\tan\text{x}-\sec\text{x}\big]^{\pi}_0$
$\Rightarrow\text{I}=(\tan\pi-\sec\pi)-(\tan0-\sec0)$
$\Rightarrow\text{I}=0+1-(0-1)$
$\Rightarrow\text{I}=1+1$
$\Rightarrow\text{I}=2$
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