Exams Joint Entrance Examination Mathematics Definite Integration Question 42 Question 42 of 246 Medium MCQ
∫ 0 π / 4 cos 2 x sin 2 x ( cos 3 x + sin 3 x ) 2 d x \int _{0}^{\pi /4}\frac{{\cos }^{2}x{\sin }^{2}x}{{({\cos }^{3}x+{\sin }^{3}x)}^{2}}dx ∫ 0 π /4 ( c o s 3 x + s i n 3 x ) 2 c o s 2 x s i n 2 x d x
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∫ 0 π / 4 cos 2 x ⋅ sin 2 x ( cos 3 x + sin 3 x ) 2 d x = ∫ 0 π / 4 tan 2 x ⋅ sec 2 x ( 1 + tan 3 x ) 2 d x \begin{gathered} \int _{0}^{\pi /4}\frac{{\cos }^{2}x\cdot {\sin }^{2}x}{{({\cos }^{3}x+{\sin }^{3}x)}^{2}}dx \\ =\int _{0}^{\pi /4}\frac{{\tan }^{2}x\cdot {\sec }^{2}x}{{(1+{\tan }^{3}x)}^{2}}dx \end{gathered} ∫ 0 π /4 ( cos 3 x + sin 3 x ) 2 cos 2 x ⋅ sin 2 x d x = ∫ 0 π /4 ( 1 + tan 3 x ) 2 tan 2 x ⋅ sec 2 x d x
Let tan x = t \tan x=t tan x = t
∫ 0 1 t 2 d t ( 1 + t 3 ) 2 \int _{0}^{1}\frac{{t}^{2}dt}{{(1+{t}^{3})}^{2}} ∫ 0 1 ( 1 + t 3 ) 2 t 2 d t
Let 1 + t 3 = z 1+{t}^{3}=z 1 + t 3 = z
3 t 2 d t = d z 1 3 ∫ 1 2 d z z 2 = 1 3 ( − 1 z ) ∣ 2 1 = − 1 3 ( 1 2 − 1 ) = 1 6 \begin{gathered} 3{t}^{2}dt=dz \\ \frac{1}{3}\int _{1}^{2}\frac{dz}{{z}^{2}}={\frac{1}{3}(-\frac{1}{z})|}_{2}^{1} \\ =-\frac{1}{3}(\frac{1}{2}-1)=\frac{1}{6} \end{gathered} 3 t 2 d t = d z 3 1 ∫ 1 2 z 2 d z = 3 1 ( − z 1 ) ∣ 2 1 = − 3 1 ( 2 1 − 1 ) = 6 1 Answer the question to unlock