Rewrite the integrand
We are given
I = ∫ sin 3 / 2 x + cos 3 / 2 x sin 3 x cos 3 x sin ( x − θ ) d x . I=\int \frac{\sin ^{3/2} x+\cos ^{3/2} x}{\sqrt{\sin ^3 x \cos ^3 x \,\sin (x-\theta)}}\,dx. I = ∫ sin 3 x cos 3 x sin ( x − θ ) sin 3/2 x + cos 3/2 x d x . Split the numerator:
sin 3 / 2 x sin 3 x cos 3 x sin ( x − θ ) = 1 cos 3 / 2 x sin ( x − θ ) , \frac{\sin ^{3/2}x}{\sqrt{\sin^3x\cos^3x\sin(x-\theta)}}
=\frac{1}{\cos^{3/2}x\sqrt{\sin(x-\theta)}}, sin 3 x cos 3 x sin ( x − θ ) sin 3/2 x = cos 3/2 x sin ( x − θ ) 1 , and similarly
cos 3 / 2 x sin 3 x cos 3 x sin ( x − θ ) = 1 sin 3 / 2 x sin ( x − θ ) . \frac{\cos ^{3/2}x}{\sqrt{\sin^3x\cos^3x\sin(x-\theta)}}
=\frac{1}{\sin^{3/2}x\sqrt{\sin(x-\theta)}}. sin 3 x cos 3 x sin ( x − θ ) cos 3/2 x = sin 3/2 x sin ( x − θ ) 1 . So
I = ∫ ( 1 cos 3 / 2 x sin ( x − θ ) + 1 sin 3 / 2 x sin ( x − θ ) ) d x . I=\int \left(\frac{1}{\cos^{3/2}x\sqrt{\sin(x-\theta)}}+\frac{1}{\sin^{3/2}x\sqrt{\sin(x-\theta)}}\right)dx. I = ∫ ( cos 3/2 x sin ( x − θ ) 1 + sin 3/2 x sin ( x − θ ) 1 ) d x . Now use
sin ( x − θ ) = sin x cos θ − cos x sin θ . \sin(x-\theta)=\sin x\cos\theta-\cos x\sin\theta. sin ( x − θ ) = sin x cos θ − cos x sin θ . Then
cos θ tan x − sin θ = sin x cos θ − cos x sin θ cos x = sin ( x − θ ) cos x , \cos\theta\tan x-\sin\theta
=\frac{\sin x\cos\theta-\cos x\sin\theta}{\cos x}
=\frac{\sin(x-\theta)}{\cos x}, cos θ tan x − sin θ = cos x sin x cos θ − cos x sin θ = cos x sin ( x − θ ) , so
cos θ tan x − sin θ = sin ( x − θ ) cos x . \sqrt{\cos\theta\tan x-\sin\theta}=\frac{\sqrt{\sin(x-\theta)}}{\sqrt{\cos x}}. cos θ tan x − sin θ = cos x sin ( x − θ ) . Also,
cos θ − sin θ cot x = sin x cos θ − cos x sin θ sin x = sin ( x − θ ) sin x , \cos\theta-\sin\theta\cot x
=\frac{\sin x\cos\theta-\cos x\sin\theta}{\sin x}
=\frac{\sin(x-\theta)}{\sin x}, cos θ − sin θ cot x = sin x sin x cos θ − cos x sin θ = sin x sin ( x − θ ) , so
cos θ − sin θ cot x = sin ( x − θ ) sin x . \sqrt{\cos\theta-\sin\theta\cot x}=\frac{\sqrt{\sin(x-\theta)}}{\sqrt{\sin x}}. cos θ − sin θ cot x = sin x sin ( x − θ ) . This strongly suggests the antiderivative is of the form
A cos θ tan x − sin θ + B cos θ − sin θ cot x + C . A\sqrt{\cos\theta\tan x-\sin\theta}+B\sqrt{\cos\theta-\sin\theta\cot x}+C. A cos θ tan x − sin θ + B cos θ − sin θ cot x + C .
Differentiate the first square-root term
Let
u = cos θ tan x − sin θ . u=\cos\theta\tan x-\sin\theta. u = cos θ tan x − sin θ . Then
d d x u = u ′ 2 u . \frac{d}{dx}\sqrt{u}=\frac{u'}{2\sqrt{u}}. d x d u = 2 u u ′ . Since
u ′ = cos θ sec 2 x , u'=\cos\theta\sec^2 x, u ′ = cos θ sec 2 x , we get
d d x cos θ tan x − sin θ = cos θ sec 2 x 2 cos θ tan x − sin θ . \frac{d}{dx}\sqrt{\cos\theta\tan x-\sin\theta}
=\frac{\cos\theta\sec^2 x}{2\sqrt{\cos\theta\tan x-\sin\theta}}. d x d cos θ tan x − sin θ = 2 cos θ tan x − sin θ cos θ sec 2 x . Using
cos θ tan x − sin θ = sin ( x − θ ) cos x , \sqrt{\cos\theta\tan x-\sin\theta}=\frac{\sqrt{\sin(x-\theta)}}{\sqrt{\cos x}}, cos θ tan x − sin θ = cos x sin ( x − θ ) , this becomes
cos θ sec 2 x 2 sin ( x − θ ) cos x = cos θ 2 ⋅ 1 cos 2 x ⋅ cos x sin ( x − θ ) = cos θ 2 cos 3 / 2 x sin ( x − θ ) . \frac{\cos\theta\sec^2 x}{2\,\frac{\sqrt{\sin(x-\theta)}}{\sqrt{\cos x}}}
=\frac{\cos\theta}{2}\cdot \frac{1}{\cos^2 x}\cdot \frac{\sqrt{\cos x}}{\sqrt{\sin(x-\theta)}}
=\frac{\cos\theta}{2\cos^{3/2}x\sqrt{\sin(x-\theta)}}. 2 c o s x s i n ( x − θ ) cos θ sec 2 x = 2 cos θ ⋅ cos 2 x 1 ⋅ sin ( x − θ ) cos x = 2 cos 3/2 x sin ( x − θ ) cos θ . So if this term is multiplied by A A A
A ⋅ cos θ 2 cos 3 / 2 x sin ( x − θ ) . A\cdot \frac{\cos\theta}{2\cos^{3/2}x\sqrt{\sin(x-\theta)}}. A ⋅ 2 cos 3/2 x sin ( x − θ ) cos θ . To match
1 cos 3 / 2 x sin ( x − θ ) , \frac{1}{\cos^{3/2}x\sqrt{\sin(x-\theta)}}, cos 3/2 x sin ( x − θ ) 1 , we need
A ⋅ cos θ 2 = 1 ⟹ A = 2 cos θ = 2 sec θ . A\cdot \frac{\cos\theta}{2}=1
\implies A=\frac{2}{\cos\theta}=2\sec\theta. A ⋅ 2 cos θ = 1 ⟹ A = cos θ 2 = 2 sec θ .
Differentiate the second square-root term
Let
v = cos θ − sin θ cot x . v=\cos\theta-\sin\theta\cot x. v = cos θ − sin θ cot x . Then
v ′ = − sin θ d d x ( cot x ) = − sin θ ( − csc 2 x ) = sin θ csc 2 x . v'= -\sin\theta\,\frac{d}{dx}(\cot x)= -\sin\theta(-\csc^2 x)=\sin\theta\csc^2 x. v ′ = − sin θ d x d ( cot x ) = − sin θ ( − csc 2 x ) = sin θ csc 2 x . Hence
d d x v = sin θ csc 2 x 2 cos θ − sin θ cot x . \frac{d}{dx}\sqrt{v}=\frac{\sin\theta\csc^2 x}{2\sqrt{\cos\theta-\sin\theta\cot x}}. d x d v = 2 cos θ − sin θ cot x sin θ csc 2 x . Using
cos θ − sin θ cot x = sin ( x − θ ) sin x , \sqrt{\cos\theta-\sin\theta\cot x}=\frac{\sqrt{\sin(x-\theta)}}{\sqrt{\sin x}}, cos θ − sin θ cot x = sin x sin ( x − θ ) , we get
sin θ csc 2 x 2 sin ( x − θ ) sin x = sin θ 2 ⋅ 1 sin 2 x ⋅ sin x sin ( x − θ ) = sin θ 2 sin 3 / 2 x sin ( x − θ ) . \frac{\sin\theta\csc^2 x}{2\,\frac{\sqrt{\sin(x-\theta)}}{\sqrt{\sin x}}}
=\frac{\sin\theta}{2}\cdot \frac{1}{\sin^2 x}\cdot \frac{\sqrt{\sin x}}{\sqrt{\sin(x-\theta)}}
=\frac{\sin\theta}{2\sin^{3/2}x\sqrt{\sin(x-\theta)}}. 2 s i n x s i n ( x − θ ) sin θ csc 2 x = 2 sin θ ⋅ sin 2 x 1 ⋅ sin ( x − θ ) sin x = 2 sin 3/2 x sin ( x − θ ) sin θ . So if this term is multiplied by B B B
B ⋅ sin θ 2 sin 3 / 2 x sin ( x − θ ) . B\cdot \frac{\sin\theta}{2\sin^{3/2}x\sqrt{\sin(x-\theta)}}. B ⋅ 2 sin 3/2 x sin ( x − θ ) sin θ . To match
1 sin 3 / 2 x sin ( x − θ ) , \frac{1}{\sin^{3/2}x\sqrt{\sin(x-\theta)}}, sin 3/2 x sin ( x − θ ) 1 , we need
B ⋅ sin θ 2 = 1 ⟹ B = 2 sin θ = 2 csc θ . B\cdot \frac{\sin\theta}{2}=1
\implies B=\frac{2}{\sin\theta}=2\csc\theta. B ⋅ 2 sin θ = 1 ⟹ B = sin θ 2 = 2 csc θ .
Compute A B AB A B
A B = ( 2 sec θ ) ( 2 csc θ ) = 4 sec θ csc θ . AB=(2\sec\theta)(2\csc\theta)=4\sec\theta\csc\theta. A B = ( 2 sec θ ) ( 2 csc θ ) = 4 sec θ csc θ . Now
sin 2 θ = 2 sin θ cos θ ⟹ csc 2 θ = 1 2 sin θ cos θ . \sin 2\theta=2\sin\theta\cos\theta
\implies \csc 2\theta=\frac{1}{2\sin\theta\cos\theta}. sin 2 θ = 2 sin θ cos θ ⟹ csc 2 θ = 2 sin θ cos θ 1 . Thus
4 sec θ csc θ = 4 sin θ cos θ = 8 csc 2 θ . 4\sec\theta\csc\theta
=\frac{4}{\sin\theta\cos\theta}
=8\csc 2\theta. 4 sec θ csc θ = sin θ cos θ 4 = 8 csc 2 θ . Therefore,
A B = 8 csc ( 2 θ ) . \boxed{AB=8\csc(2\theta)}. A B = 8 csc ( 2 θ ) . So the correct option is B .
Comparison with stored answer
Stored correct answer: B
Our derived answer: B
They agree.