Exams Joint Entrance Examination Mathematics Definite Integration Question 234 Question 234 of 246 Medium MCQ
If I = ∫ 0 π 2 sin 3 2 x sin 3 2 x + cos 3 2 x d x \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} \mathrm{~d} x I = ∫ 0 2 π s i n 2 3 x + c o s 2 3 x s i n 2 3 x d x , then ∫ 0 2 I x sin x cos x sin 4 x + cos 4 x d x \int_0^{2I} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} \mathrm{~d} x ∫ 0 2 I s i n 4 x + c o s 4 x x s i n x c o s x d x equals :
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For I
Apply king ( P − 5 \mathrm{P}-5 P − 5 ) and add
\[ 2 I = ∫ 0 π / 2 d x = π 2 ⇒ I = π 4 I 2 = ∫ 0 π / 2 x sin x cos x sin 4 x + cos 4 x d x \begin{aligned}
& 2 I=\int_0^{\pi / 2} d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4} \\
& I_2=\int_0^{\pi / 2} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x
\end{aligned} 2 I = ∫ 0 π /2 d x = 2 π ⇒ I = 4 π I 2 = ∫ 0 π /2 sin 4 x + cos 4 x x sin x cos x d x \]
Apply king and add
\[ I 2 = π 4 ∫ 0 π / 2 tan x s e c 2 x d x tan 4 x + 1 put tan 2 x = t π 8 ∫ 0 ∞ d t t 2 + 1 = π 8 ⋅ π 2 = π 2 16 \begin{aligned}
& \mathrm{I}_2=\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan \mathrm{xsec}{ }^2 \mathrm{xdx}}{\tan ^4 \mathrm{x}+1} \\
& \text { put } \tan ^2 \mathrm{x}=\mathrm{t} \\
& \frac{\pi}{8} \int_0^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^2+1} \\
& =\frac{\pi}{8} \cdot \frac{\pi}{2}=\frac{\pi^2}{16}
\end{aligned} I 2 = 4 π ∫ 0 π /2 tan 4 x + 1 tan xsec 2 xdx put tan 2 x = t 8 π ∫ 0 ∞ t 2 + 1 dt = 8 π ⋅ 2 π = 16 π 2 \] Answer the question to unlock