Mathematics

# Prove:$\int _ { 0 } ^ { \pi } \dfrac { x \sin x } { 1 + \cos ^ { 2 } x } d x = \dfrac { \pi ^ { 2 } } { 4 }$

##### SOLUTION
$I = \int_{0}^{\pi} \dfrac {x \, sin \,x}{1+ cos^2x}dx$...(1)
using property $\infty$ definite integral

$\int_{a}^{b} f(x)dx =\int_{a}^{b} f(a+b-x)dx$

$I = \int_{a}^{b}\dfrac {(\pi-x)sin (\pi-x)}{1+cos^2(\pi-x)}dx$

$I = \int_{a}^{b}\dfrac {(\pi-x)sinx}{1+cos^2x}dx$....(2)

adding $(1)$& $(2)$, we get

$I+I=\int_{a}^{b}\dfrac {(\pi-x)sinx}{1+cos^2x}dx + \int_{a}^{b}\dfrac {(\pi-x)sinx}{1+cos^2x}$

$2I = \int_{a}^{b}\dfrac {(\pi- Sin x)sinx}{1+cos^2x}dx$

$I =\dfrac {\pi}{2}\int_{0}^{\pi} \dfrac {Sin x}{1+Cos^2x}dx$

$=\dfrac {\pi}{2}\cdot 2 \int_{0}^{\pi/2} \dfrac {Sin x}{1+Cos^2x}dx$

$= -\pi \int_{0}^{\pi/2} \frac {d(cos\,x)}{1+Cos^2x}dx$

$= - \pi [tan^{-1}(cos\,x)]_0^{\pi/2}$

$-\pi [ tan^{-1} cos \dfrac{\pi}2-tan^{-1}(cos \,0)]$

$-\pi [ 0- \dfrac{\pi} 4]$

$=\dfrac{\pi^2}4$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Hard
Repeated application of integration by parts gives us the reduction formula, if the integrand is dependent on a natural number $n$.

If $\displaystyle \int \frac {dx}{x^n \sqrt {ax + b}} = -\frac {\sqrt {ax + b}}{(n - 1) bx^{n - 1}} - A \int \frac {dx}{x^{n - 1} \sqrt {ax + b}} + C$, then $A$ is equal to
• A. $\displaystyle \frac {2n - 3}{2n - 1}$
• B. $\displaystyle \frac {n - 2}{n - 1} \frac {a}{b}$
• C. $\displaystyle \frac {n - 3}{n - 2} \frac {a}{b}$
• D. $\displaystyle \frac {2n - 3}{2n - 2} \frac {a}{b}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The integral $\displaystyle \int{\frac{dx}{(x+1)^{ \frac{3}{4}} (x-2)^{\frac{5}{4}}}}$ is equal to
• A. $\displaystyle \frac{-4}{3} \left( \frac{x-2}{x+1} \right)^{\displaystyle \frac{1}{4}} +C$
• B. $\displaystyle 4 \left( \frac{x+1}{x-2} \right)^{\displaystyle \frac{1}{4}} +C$
• C. $\displaystyle 4 \left( \frac{x-2}{x+1} \right)^{\displaystyle \frac{1}{4}} +C$
• D. $\displaystyle \frac{-4}{3} \left( \frac{x+1}{x-2} \right)^{\displaystyle \frac{1}{4}} +C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Find the value of $\displaystyle\int e^x(\tan x-\log|\cos x|)dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle \int_{1}^{2}\mathrm{x}^{2\mathrm{x}}[1+\log \mathrm{x}]\mathrm{d}\mathrm{x}=$
• A. $\displaystyle \frac{9}{2}$
• B. $\displaystyle \frac{11}{2}$
• C. $\displaystyle \frac{13}{2}$
• D. $\displaystyle \frac{15}{2}$

$_{0}^{\pi/2}\int x \sin x dx$.