Mathematics

# Evaluate: $\displaystyle\int_{0}^{\pi}\dfrac{x\sin x}{1+\cos^{2}x}dx$

##### SOLUTION
$\begin{array}{l}\text { Let } I=\displaystyle \int_{0}^{\pi} \dfrac{x \sin x}{1+\cos ^{2} x} \mathrm{d} \mathrm{x} \quad \Rightarrow \quad I=\displaystyle \int_{0}^{\pi} \dfrac{(\pi-x) \sin (\pi-x) \mathrm{d} x}{1+\cos ^{2}(\pi-x)} \\=\displaystyle \int_{0}^{\pi} \dfrac{(\pi-x) \sin x \mathrm{d} x}{1+\cos ^{2} x}=\pi \int_{0}^{\pi} \dfrac{\sin x \mathrm{d} x}{1+\cos ^{2} x}-I \\\text { or } \displaystyle 2 I=\pi \int_{0}^{\pi} \dfrac{\sin x \mathrm{dx}}{1+\cos ^{2} x} \text { or } I=\dfrac{\pi}{2} \int_{0}^{\pi} \dfrac{\sin x \mathrm{dx}}{1+\cos ^{2} x}\end{array}\\$

Put $\cos x=t$ so that $-\sin x d x=d t$

The limits are, when $x=0, t=1$ and $x=\pi, t=-1,$ we get

$\begin{array}{l}I=\dfrac{-\pi}{2} \displaystyle \int_{1}^{-1} \dfrac{\mathrm{dt}}{1+t^{2}}=\pi \int_{0}^{1} \dfrac{\mathrm{dt}}{1+t^{2}}\\ \left[\because \int_{a}^{-a} f(x) \mathrm{d} \mathrm{x}=-\int_{-a}^{a} f(x) \mathrm{d} \mathrm{x} \text { and } = -2\int_{0}^{2 a} f(x) \mathrm{d} \mathrm{x} ,\ \ \text{if f(x) is even function}\right]. \\I=\pi | \tan ^{-1} t |_0^1=\pi \left[ \tan ^{-1} 1-\tan ^{-1} 0\right]=\pi\left[\dfrac{\pi}{4}-0\right]=\dfrac{\pi^{2}}{4}\end{array}\\$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
If $g(1)=g(2)$ then the value of  $\int _{ 1 }^{ 2 }{ { \left[ f\{ g(x)\} \right] }^{ -1 } } f'\{ g(x)\} g'(x)dx\quad is$
• A. $1$
• B. $2$
• C. none of these
• D. $0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int \frac{1}{(x^{6}-1)}dx$
• A. 1/2($\frac { 1 }{ 6 } \ln { \frac { { \left( x+1 \right) }^{ 2 } }{ { x }^{ 2 }-x+1 } } -\frac { 1 }{ \sqrt { 3 } } \arctan { \frac { 2x-1 }{ \sqrt { 3 } } -\frac { 1 }{ 6 } \ln { \frac { { \left( x-1 \right) }^{ 2 } }{ { x }^{ 2 }+x+1 } } } -\frac { 1 }{ \sqrt { 3 } } \arctan { \frac { 2x+1 }{ \sqrt { 3 } } }$)+k
• B. 1/2($\frac { 1 }{ 6 } \ln { \frac { { \left( x-1 \right) }^{ 2 } }{ { x }^{ 2 }+x+1 } } -\frac { 1 }{ \sqrt { 3 } } arccot { \frac { 2x+1 }{ \sqrt { 3 } } +\frac { 1 }{ 6 } \ln { \frac { { \left( x+1 \right) }^{ 2 } }{ { x }^{ 2 }-x+1 } } } +\frac { 1 }{ \sqrt { 3 } } arccot { \frac { 2x-1 }{ \sqrt { 3 } } }$)+k
• C. 1/2($\frac { 1 }{ 6 } \ln { \frac { { \left( x-1 \right) }^{ 2 } }{ { x }^{ 2 }+x+1 } } -\frac { 1 }{ \sqrt { 3 } } arccot { \frac { 2x+1 }{ \sqrt { 3 } } -\frac { 1 }{ 6 } \ln { \frac { { \left( x+1 \right) }^{ 2 } }{ { x }^{ 2 }-x+1 } } } -\frac { 1 }{ \sqrt { 3 } } arccot { \frac { 2x-1 }{ \sqrt { 3 } } }$)+k
• D. 1/2($\frac { 1 }{ 6 } \ln { \frac { { \left( x-1 \right) }^{ 2 } }{ { x }^{ 2 }+x+1 } } -\frac { 1 }{ \sqrt { 3 } } \arctan { \frac { 2x+1 }{ \sqrt { 3 } } -\frac { 1 }{ 6 } \ln { \frac { { \left( x+1 \right) }^{ 2 } }{ { x }^{ 2 }-x+1 } } } -\frac { 1 }{ \sqrt { 3 } } \arctan { \frac { 2x-1 }{ \sqrt { 3 } } }$)+k

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
The value of $\displaystyle \int_{1/3}^{3}\dfrac {\tan \left (x^{2} + \dfrac {1}{x^{2}}\right )}{\sin \left (x + \dfrac {1}{x}\right )} \dfrac {dx}{x}$ is

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $\displaystyle\int \dfrac{dx}{x^3(1+x^6)^{2/3}}=x$. $f(x)\cdot (1+x^6)^{1/3}+C$, then $f(x)$ is equal to?
• A. $\dfrac{-1}{2x^2}$
• B. $\dfrac{-1}{6x^2}$
• C. $\dfrac{1}{6x^2}$
• D. $\dfrac{-1}{2x^3}$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$