JEE Main 2025 29 January Morning Shift
Exam JEE Main
Year 2025
Date 29 January
Shift Morning Shift
Subject Mathematics
Chapter Definite Integration JEE Main 2025 · 29 January · Morning Shift · Mathematics · Definite Integration
The integral (80 ∫_0^ (π)/(4) (( θ+ θ)/(9+16 2 θ)) d θ) is equal to...
The integral (80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16 \sin 2 \theta}\right) d \theta) is equal to :
Options A (3 \log _e 4) B (4 \log _{\mathrm{e}} 3) C (6 \log _e 4) D (2 \log _e 3) Correct answer B. (4 \log _{\mathrm{e}} 3)
Step-by-step solution
I = ∫ 0 π 4 ( sin θ + cos θ 9 − 16 sin 2 θ ) d θ Take sin θ − cos θ = t ( cos θ + sin θ ) d θ = d t ( sin θ − cos θ ) 2 = t 2 ⇒ sin 2 θ = 1 − t 2 θ = 0 → t = − 1 θ = π 4 → t = 0 I = ∫ − 1 0 d t 9 + 16 ( 1 − t 2 ) = 1 16 ∫ − 1 0 d t 25 16 16 − t 2 = 1 4 [ 1 10 log ∣ 5 + 4 t 5 − 4 t ] − 1 0 = 1 40 [ 0 + log e 9 ] I = log e 9 40 80 I = 2 log e 9 80 I = 4 log e 3 \begin{aligned} & I=\int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9-16 \sin 2 \theta}\right) d \theta \\ & \text{ Take } \sin \theta-\cos \theta=t \\ & (\cos \theta+\sin \theta) d \theta=d t \\ & (\sin \theta-\cos \theta)^2=t^2 \\ & \Rightarrow \sin 2 \theta=1-t^2 \\ & \theta=0 \rightarrow t=-1 \\ & \quad \theta=\frac{\pi}{4} \rightarrow t=0 \\ & I=\int_{-1}^0 \frac{d t}{9+16\left(1-t^2\right)} \\ & =\frac{1}{16} \int_{-1}^0 \frac{d t}{25} \frac{16}{16}-t^2 \\ & =\frac{1}{4}\left[\frac{1}{10} \log _{\left\lvert\, \frac{5+4 t}{}\right.}^{5-4 t}\right]_{-1}^0 \\ & =\frac{1}{40}\left[0+\log _e 9\right] \\ & I=\frac{\log _e 9}{40} \\ & 80 I=2 \log _e 9 \\ & 80 I=4 \log _e 3\end{aligned} I = ∫ 0 4 π ( 9 − 16 sin 2 θ sin θ + cos θ ) d θ Take sin θ − cos θ = t ( cos θ + sin θ ) d θ = d t ( sin θ − cos θ ) 2 = t 2 ⇒ sin 2 θ = 1 − t 2 θ = 0 → t = − 1 θ = 4 π → t = 0 I = ∫ − 1 0 9 + 16 ( 1 − t 2 ) d t = 16 1 ∫ − 1 0 25 d t 16 16 − t 2 = 4 1 [ 10 1 log ∣ 5 + 4 t 5 − 4 t ] − 1 0 = 40 1 [ 0 + log e 9 ] I = 40 log e 9 80 I = 2 log e 9 80 I = 4 log e 3
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