Exams Joint Entrance Examination Mathematics Definite Integration Question 35 Question 35 of 246 Medium MCQ
If I = ∫ π 2 0 sin 3 2 x sin 3 2 x + cos 3 2 x d x I={\int }_{\frac{\pi }{2}}^{0}\frac{{\sin }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx I = ∫ 2 π 0 s i n 2 3 x + c o s 2 3 x s i n 2 3 x d x , then ∫ 2 I 0 x sin x cos x sin 4 x + cos 4 x d x {\int }_{2I}^{0}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx ∫ 2 I 0 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 P-5 P − 5 ) and add
2 I = ∫ π / 2 0 d x = π 2 ⇒ I = π 4 I 2 = ∫ π / 2 0 x sin x cos x sin 4 x + cos 4 x d x \begin{gathered} 2I={\int }_{\pi /2}^{0}dx=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4} \\ {I}_{2}={\int }_{\pi /2}^{0}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx \end{gathered} 2 I = ∫ π /2 0 d x = 2 π ⇒ I = 4 π I 2 = ∫ π /2 0 sin 4 x + cos 4 x x sin x cos x d x
Apply king and add
I 2 = π 4 ∫ π / 2 0 tan x s e c 2 x d x tan 4 x + 1 tan 2 x = t π 8 ∫ ∞ 0 d t t 2 + 1 = π 8 ⋅ π 2 = π 2 16 \begin{gathered} {I}_{2}=\frac{\pi }{4}{\int }_{\pi /2}^{0}\frac{\tan xsec{}^{2}xdx}{{\tan }^{4}x+1} \\ {\tan }^{2}x=t \\ \frac{\pi }{8}{\int }_{\infty}^{0}\frac{dt}{{t}^{2}+1} \\ =\frac{\pi }{8}\cdot \frac{\pi }{2}=\frac{{\pi }^{2}}{16} \end{gathered} I 2 = 4 π ∫ π /2 0 tan 4 x + 1 tan x sec 2 x d x tan 2 x = t 8 π ∫ ∞ 0 t 2 + 1 d t = 8 π ⋅ 2 π = 16 π 2 Answer the question to unlock