JEE Main 2024 4 April Morning Shift
Exam JEE Main
Year 2024
Date 4 April
Shift Morning Shift
Subject Mathematics
Chapter Definite Integration JEE Main 2024 · 4 April · Morning Shift · Mathematics · Definite Integration
If ∫_0^ (π)/(4) ( ^2 x)/(1+ x x) ~d x= 1 a _ e ( a 3 )+ π b 3 ,...
If
∫ 0 π 4 sin 2 x 1 + sin x cos x d x = 1 a log e ( a 3 ) + π b 3 \int_0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} \mathrm{~d} x=\frac{1}{\mathrm{a}} \log _{\mathrm{e}}\left(\frac{\mathrm{a}}{3}\right)+\frac{\pi}{\mathrm{b} \sqrt{3}} ∫ 0 4 π 1 + s i n x c o s x s i n 2 x d x = a 1 log e ( 3 a ) + b 3 π , where
a , b ∈ N \mathrm{a}, \mathrm{b} \in \mathbf{N} a , b ∈ N , then
a + b \mathrm{a}+\mathrm{b} a + b is equal to_________
Step-by-step solution
∫ 0 π 2 sin 2 x 1 + 1 2 sin 2 x d x = ∫ 0 π 4 1 − cos 2 x 2 + sin 2 x d x ∫ 1 2 + sin 2 x − ∫ cos 2 x 2 + sin 2 x \begin{aligned}
& \int_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+\frac{1}{2} \sin 2 x} d x=\int_0^{\frac{\pi}{4}} \frac{1-\cos 2 x}{2+\sin 2 x} d x \\
& \int \frac{1}{2+\sin 2 x}-\int \frac{\cos 2 x}{2+\sin 2 x}
\end{aligned} ∫ 0 2 π 1 + 2 1 sin 2 x sin 2 x d x = ∫ 0 4 π 2 + sin 2 x 1 − cos 2 x d x ∫ 2 + sin 2 x 1 − ∫ 2 + sin 2 x cos 2 x \left( \mathrm{I}
_1\right) \quad-\quad\left( \mathrm{I}
_2\right) ( I 1 ) = ∫ d x 2 + 2 tan x 1 + tan 2 x ∫ 0 π 4 sec 2 x d x 2 tan 2 x + 2 tan x + 2 \begin{aligned}
& \left(\mathrm{I}_1\right)=\int \frac{\mathrm{dx}}{2+\frac{2 \tan \mathrm{x}}{1+\tan ^2 x}} \\
& \int_0^{\frac{\pi}{4}} \frac{\sec ^2 \mathrm{xdx}}{2 \tan ^2 \mathrm{x}+2 \tan x+2}
\end{aligned} ( I 1 ) = ∫ 2 + 1 + t a n 2 x 2 t a n x dx ∫ 0 4 π 2 tan 2 x + 2 tan x + 2 sec 2 xdx tan x = t 1 2 ∫ 0 1 d t ( t + 1 2 ) 2 + 3 4 = π 6 3 I 2 = ∫ 0 π / 4 cos 2 x 2 + sin 2 x d x = 1 2 ( ln 3 2 ) I 1 − I 2 = 1 3 π 6 + 1 2 ln 2 3 ⇒ a = 2 , b = 6 \begin{aligned}
& \tan x=t \\
& \frac{1}{2} \int_0^1 \frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{\pi}{6 \sqrt{3}} \\
& \mathrm{I}_2=\int_0^{\pi / 4} \frac{\cos 2 \mathrm{x}}{2+\sin 2 \mathrm{x}} \mathrm{dx}=\frac{1}{2}\left(\ln \frac{3}{2}\right) \\
& \mathrm{I}_1-\mathrm{I}_2=\frac{1}{\sqrt{3}} \frac{\pi}{6}+\frac{1}{2} \ln \frac{2}{3} \\
& \Rightarrow \mathrm{a}=2, \mathrm{~b}=6
\end{aligned} tan x = t 2 1 ∫ 0 1 ( t + 2 1 ) 2 + 4 3 dt = 6 3 π I 2 = ∫ 0 π /4 2 + sin 2 x cos 2 x dx = 2 1 ( ln 2 3 ) I 1 − I 2 = 3 1 6 π + 2 1 ln 3 2 ⇒ a = 2 , b = 6 Ans. 8
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