Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 15 Question 15 of 416 Hard MCQ
If ∫ sin 3 2 x + cos 3 2 x sin 3 x cos 3 x sin ( x − θ ) d x = A cos θ tan x − sin θ + B cos θ − sin θ cot x + C \int \frac{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}{\sqrt{{\sin }^{3}x{\cos }^{3}x\sin (x-\theta )}}dx=A\sqrt{\cos \theta \tan x-\sin \theta }+B\sqrt{\cos \theta -\sin \theta \cot x}+C ∫ s i n 3 x c o s 3 x s i n ( x − θ ) s i n 2 3 x + c o s 2 3 x d x = A cos θ tan x − sin θ + B cos θ − sin θ cot x + C , where C C C is the integration constant, then A B AB A B is equal to
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∫ sin 3 2 x + cos 3 2 x sin 3 x cos 3 x sin ( x − θ ) d x I = ∫ sin 3 2 x + cos 3 2 x sin 3 x cos 3 x ( sin x cos θ − cos x sin θ ) d x = ∫ sin 3 2 x sin 3 2 x cos 2 x tan x cos θ − sin θ d x + ∫ cos 3 2 x sin 2 x cos 3 2 x cos θ − cot x sin θ d x = ∫ sec 2 x tan x cos θ − sin θ d x + ∫ csc 2 x cos θ − cot x sin θ d x \begin{gathered} \int \frac{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}{\sqrt{{\sin }^{3}x{\cos }^{3}x\sin (x-\theta )}}dx \\ I=\int \frac{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}{\sqrt{{\sin }^{3}x{\cos }^{3}x(\sin x\cos \theta -\cos x\sin \theta )}}dx \\ =\int \frac{{\sin }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x{\cos }^{2}x\sqrt{\tan x\cos \theta -\sin \theta }}dx+\int \frac{{\cos }^{\frac{3}{2}}x}{{\sin }^{2}x{\cos }^{\frac{3}{2}}x\sqrt{\cos \theta -\cot x\sin \theta }}dx= \\ \int \frac{{\sec }^{2}x}{\sqrt{\tan x\cos \theta -\sin \theta }}dx+\int \frac{{\csc }^{2}x}{\sqrt{\cos \theta -\cot x\sin \theta }}dx \end{gathered} ∫ sin 3 x cos 3 x sin ( x − θ ) sin 2 3 x + cos 2 3 x d x I = ∫ sin 3 x cos 3 x ( sin x cos θ − cos x sin θ ) sin 2 3 x + cos 2 3 x d x = ∫ sin 2 3 x cos 2 x tan x cos θ − sin θ sin 2 3 x d x + ∫ sin 2 x cos 2 3 x cos θ − cot x sin θ cos 2 3 x d x = ∫ tan x cos θ − sin θ sec 2 x d x + ∫ cos θ − cot x sin θ csc 2 x d x
I = I 1 + I 2 I={I}_{1}+{I}_{2} I = I 1 + I 2 ...... {Let}
For I 1 {I}_{1} I 1 , let tan x cos θ − sin θ = t 2 \tan x\cos \theta -\sin \theta ={t}^{2} tan x cos θ − sin θ = t 2
sec 2 x d x = 2 t d t cos θ {\sec }^{2}xdx=\frac{2tdt}{\cos \theta } sec 2 x d x = c o s θ 2 t d t
For I 2 {I}_{2} I 2 , let cos θ − cot x sin θ = z 2 \cos \theta -\cot x\sin \theta ={z}^{2} cos θ − cot x sin θ = z 2
csc 2 x d x = 2 z d z sin θ {\csc }^{2}xdx=\frac{2zdz}{\sin \theta } csc 2 x d x = s i n θ 2 z d z
I = I 1 + I 2 = ∫ 2 t d t cos θ t + ∫ 2 z d z sin θ z = 2 t cos θ + 2 z sin θ = 2 sec θ tan x cos θ − sin θ + 2 csc θ cos θ − cot x sin θ A B = 8 csc 2 θ \begin{gathered} I={I}_{1}+{I}_{2} \\ =\int \frac{2tdt}{\cos \theta t}+\int \frac{2zdz}{\sin \theta z} \\ =\frac{2t}{\cos \theta }+\frac{2z}{\sin \theta } \\ =2\sec \theta \sqrt{\tan x\cos \theta -\sin \theta }+2\csc \theta \sqrt{\cos \theta -\cot x\sin \theta } \\ \, AB=8\csc 2\theta \end{gathered} I = I 1 + I 2 = ∫ cos θ t 2 t d t + ∫ sin θ z 2 z d z = cos θ 2 t + sin θ 2 z = 2 sec θ tan x cos θ − sin θ + 2 csc θ cos θ − cot x sin θ A B = 8 csc 2 θ Answer the question to unlock