Mathematics

# The value of $\int\limits_0^{\pi /2} {\dfrac{{x\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx\,is}$ is

$\dfrac{{{\pi ^2}}}{{16}}$

##### SOLUTION
$\displaystyle I = \int_{0}^{\pi/2} \frac{x sinx \, cosx}{sin^4x + cos^4 x}dx$
using $\displaystyle \int_{0}^{a} f(n)dx = \int_{0}^{a} f(a-x) dx,$ we get
$\displaystyle I= \int_{0}^{\pi/2} \frac{(\pi/2 -\pi) sin (\pi /2 -\pi) cos (\pi/2 - \pi)}{sin^4 (\pi/2 -\pi)+ cos^4 (\pi /2 -\pi)}$
$\displaystyle = \int_{0}^{\pi/2} \frac{(\pi/2-\pi)cosx sinx}{cos^4x + sin^4 x} dx$
$\displaystyle I = \int_{0}^{\pi/2} \frac{cos sinx}{cos^4 x+ sin^4 x}dx - \int_{0}^{\pi/2} \frac{+xcosx \, sinx}{cos^4x+sin^4x}dx$
$\displaystyle \therefore 2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{sinx cosx}{sin^4 x + cos^4x} dx - I$
$\displaystyle \therefore I = \dfrac{\pi}{2}\int_{0}^{1} \frac{dt/2}{t^2+(1-t)^2}$
as $x = \pi / 2,$ then t = 1
and $x = 0 ,$ then $t =0$
$\displaystyle = \frac{\pi}{8} \int_{0}^{1} \dfrac{dt}{t^2+t^2+1-2t}$
$\displaystyle = \frac{\pi}{8} \int_{0}^{1} \dfrac{dt}{2t^2-2t+1}$
$\displaystyle = \dfrac{\pi}{16} \int_{0}^{1} \dfrac{dt}{(t-\dfrac{1}{2})^2 + (\dfrac{1}{2})^2}$
$\displaystyle = \dfrac{\pi}{16} \times \frac{1}{1/2} [ tan^{-1} (\frac{t-1/2}{1/2})]$
$\displaystyle \because \int \frac{1}{x^2+a^2}dx = \frac{1}{a} tan^{-1}x/a$
$\displaystyle = [tan^{-1} 2(1-1/2)-tan^{-1}2(0-1/2)]$
$\displaystyle = \dfrac{\pi}{8} [tan^{-1}1-tan^{-1}(-1)]$
$\displaystyle \because tan^{-1}(-1) = -tan^{-1}1 = -\pi/4$
$\displaystyle I = \frac{\pi}{8} [ \frac{\pi}{4}+\frac{\pi}{4}] = \frac{\pi^2}{16}$

Its FREE, you're just one step away

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

#### Realted Questions

Q1 Single Correct Medium
let $f$ be integrable over $[0,a]$ for any real $a$. If any real $a$. If we define $I_1=\displaystyle \int_0^{\pi/2}\cos \theta f(\sin\theta +\cos^2\theta)d\theta$ and $I_2=\displaystyle \int_0^{\pi/2}\sin\theta f(\sin\theta+cos^2\theta)d\theta$,then
• A. $I_1=-I_2$
• B. $I_1=2I_2$
• C. $I_1=-2I_2$
• D. $I_1=I_2$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
$\displaystyle \int_{0}^{\pi }\frac{x dx}{1+\cos ^{2}x}.$
• A. $\dfrac{\pi}{2\sqrt{\left ( 2 \right )}}.$
• B. $\pi { \sqrt{\left ( 2 \right )}}.$
• C. $\dfrac{\pi ^{2}}{\sqrt{\left ( 2 \right )}}.$
• D. $\dfrac{\pi ^{2}}{2\sqrt{\left ( 2 \right )}}.$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate: $\displaystyle\int _ { 0 } ^ { \pi / 2 } \dfrac { \sin x \cos x } { \cos ^ { 2 } x + 3 \cos x + 2 }dx$
• A. $\ln\left(\dfrac {5}{3}\right)$
• B. $\ln\left(\dfrac {1}{3}\right)$
• C. None of these
• D. $\ln\left(\dfrac {4}{3}\right)$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\int [1 + \tan x \cdot \tan (x + \alpha)]dx$ is equal to
• A. $\cos \alpha\cdot \log |\dfrac{\sin x}{\sin (x + \alpha)}| + C$
• B. $\tan \alpha\cdot \log |\dfrac{\sin x}{\sin (x + \alpha)}| + C$
• C. $\cot \alpha\cdot \log |\dfrac{\cos (x + \alpha)}{\cos x}| + C$
• D. $\cot \alpha\cdot \log|\dfrac{\sec (x + \alpha)}{\sec x}| + C$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$