Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The integral $\int_{0}^{\frac{1}{2}}\frac{ln(1+2x)}{1+4x^{2}}dx,$ equals :
Option: 1
$\frac{\pi }{4}ln2$

Option: 2
$\frac{\pi }{8}ln2$

Option: 3
$\frac{\pi }{16}ln2$

Option: 4
$\frac{\pi }{32}ln2$

Answers (1)

best_answer

As we learnt in concept

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

- wherein

Since $\int f(x)dx=\int f(t)dt=\int f(\theta )d\theta$ all variables must be converted into single variable , $\left ( t\, or\ \theta \right )$

$I= \int_{0}^{1}\frac{In(1+2x)}{1+4x^2} dx$

Let 2x = t and dx = dt/2

$I= \frac{1}{2}\int_{0}^{1}\frac{In(1+t)}{1+t^2} dt$

$Let \: t=\tan\theta \:$
$dt=\sec^2\theta d\theta$

$I=\frac{1}{2}\int_{0}^{\pi/4}\frac{In(1+tan\theta)}{\sec^2\theta}\sec^2\theta d\theta$

$I= \frac{1}{2}\int_{0}^{\pi/4}In(1+\tan\theta)d\theta$ --------------------------------- (1)

Substituting
$y=0+\frac{\pi}{4}-x, we \:\:get$

$I= \frac{1}{2}\int_{0}^{\pi/4}In\left [ 1+tan(\frac{\pi}{4}-y) \right ] dy$

$I= \frac{1}{2}\int_{0}^{\pi/4}In\left [ 1+\frac{1-\tan y}{1+\tan y}\right ] dy$

$=\frac{1}{2}\int_{0}^{\pi/4}In \left ( \frac{2}{1+\tan y} \right )dy$

$=\frac{1}{2}\int_{0}^{\pi/4}In2\:- \frac{1}{2}\int_{0}^{\pi/4}In\: (1+\tan\theta )d\theta$ ------------------------------------ (2)

From (1) & (2),

$2I= \frac{1}{2}\int_{0}^{\pi/4}In2\:dx$

$I= \frac{1}{4}(In \:2)\frac{\pi}{4}= \frac{\pi}{16}In\:2$

Posted by

Pankaj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82⁰C. If Na₂SO₄ is 81.5% ionised, the value of x (K

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO₃ to give fraction A. The leftover organic phase was extracted with d

Trending Articles/News

anam-khan

JEE Main 2026 Paper 2 Toppers: 4 score 100 percentile; most from Kerala, one female topper

anam-khan

JEE Mains 2026 paper 2 result out on jeemain.nta.nic.in

anam-khan

JEE Mains 2026 session 2 B.Arch, B.Planning final answer key out, no question dropped

Latest Question