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NCERT Class XII Mathematics Chapter - - Solutions

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..... Miscellaneous Exercise .....
Integrate the functions in Exercises 229 to 252.
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Question : 246 of 271
Marks: +1, -0
1sin⁡3xsin⁡(x+α)\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}
Solution:  
Let I=∫1sin⁡3xsin⁡(x+α)dxI = \int \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} dx
=∫sin⁡xsin⁡3xsin⁡(x+α)dx= \int \sqrt{ \frac{\sin x}{\sin^3 x \sin(x+\alpha)} } dx
=∫1sin⁡2xsin⁡xsin⁡(x+α)dx= \int \frac{1}{\sin^2 x} \sqrt{ \frac{\sin x}{\sin(x+\alpha)} } dx
Let sin⁡(x+α)sin⁡x=t\frac{\sin(x+\alpha)}{\sin x} = t
⇒sin⁡xcos⁡(x+α)−cos⁡xsin⁡(x+α)sin⁡2xdx=dt\Rightarrow \frac{\sin x \cos(x+\alpha) - \cos x \sin(x+\alpha)}{\sin^2 x} dx = dt
⇒sin⁡[x−(x+α)]sin⁡2xdx=dt\Rightarrow \frac{\sin[x-(x+\alpha)]}{\sin^2 x} dx = dt
⇒−sin⁡αsin⁡2xdx=dt\Rightarrow -\frac{\sin\alpha}{\sin^2 x} dx = dt
∴I=∫−1sin⁡α⋅1tdt\therefore I = \int -\frac{1}{\sin\alpha} \cdot \frac{1}{\sqrt{t}} dt
=−1sin⁡α⋅t12+C= -\frac{1}{\sin\alpha} \cdot \frac{\sqrt{t}}{\frac{1}{2}} + C
=−2sin⁡αt+C= -\frac{2}{\sin\alpha} \sqrt{t} + C
=−2sin⁡αsin⁡(x+α)sin⁡x+C= -\frac{2}{\sin\alpha} \sqrt{ \frac{\sin(x+\alpha)}{\sin x} } + C
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