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  • Integrate the functions in Exercises 1 to 24. 18.

Q: Integrate the function $\frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}$.

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Given, $\frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}$

Let $I=\int \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}$

We know the identity that

$\sin (A+B) = \sin A \cos B + \cos A \sin B$

$\therefore \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}=\frac{1}{\sqrt{\sin ^3 x(\sin x \cos \alpha+\cos x \sin \alpha)}}$

$=\frac{1}{\sqrt{\sin ^3 x \cdot \sin x(\cos \alpha+\cot x \sin \alpha)}}=\frac{1}{\sqrt{\sin ^4 x(\cos \alpha+\cot x \sin \alpha)}}$

$\frac{\operatorname{cosec}^2 x}{\sqrt{(\cos \alpha+\cot x \sin \alpha)}}$

Now, let $(\cos \alpha+\cot \mathrm{x} \sin \alpha)=\mathrm{t} \Rightarrow-\operatorname{cosec}^2 \mathrm{x} \cdot \sin \alpha \mathrm{dx}=\mathrm{dt}

$\Rightarrow \int \frac{1}{\sqrt{\sin ^3 \mathrm{x} \sin (\mathrm{x}+\alpha)}} \mathrm{dx}=\int \frac{\operatorname{cosec}^2 \mathrm{x}}{\sqrt{(\cos \alpha+\cot \mathrm{x} \sin \alpha)}} \mathrm{dx}$

$=\int \frac{1}{\sqrt{\mathrm{t}}} \cdot-\frac{\mathrm{dt}}{\sin \alpha}$

$=-\frac{1}{\sin \alpha} \int \frac{1}{\sqrt{\mathrm{t}}} \cdot \mathrm{dt}$

$=-\frac{1}{\sin \alpha} \int \mathrm{t}^{-\frac{1}{2}} \cdot \mathrm{dt}$

$=-\frac{1}{\sin \alpha}\left[\frac{\mathrm{t}^{\frac{1}{2}}}{\frac{1}{2}}\right]+\mathrm{C}$

$=-\frac{2}{\sin \alpha}[\sqrt{\mathrm{t}}]+\mathrm{C}$

$=-\frac{2}{\sin \alpha}[\sqrt{(\cos \alpha+\cot \mathrm{x} \sin \alpha)}]+\mathrm{C}$

$=-\frac{2}{\sin \alpha}\left[\sqrt{\left(\cos \alpha+\frac{\cos \mathrm{x}}{\sin \mathrm{x}} \sin \alpha\right)}\right]+\mathrm{C}\end{aligned}$

$=-\frac{2}{\sin \alpha}\left[\sqrt{\frac{(\cos \alpha \sin x+\cos x \sin \alpha)}{\sin x}}\right]+C$

$\Rightarrow I=-\frac{2}{\sin \alpha}\left[\sqrt{\frac{\sin (x+\alpha)}{\sin x}}\right]+C\end{aligned}$

Posted by

HARSH KANKARIA

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Given, $\frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}$

Let $I=\int \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}} d x$

We know 

$\sin (A+B) = \sin A \cos B + \cos A \sin B$

$\therefore \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}=\frac{1}{\sqrt{\sin ^3 x(\sin x \cos \alpha+\cos x \sin \alpha)}}$

$=\frac{1}{\sqrt{\sin ^3 x \cdot \sin x(\cos \alpha+\cot x \sin \alpha)}}=\frac{1}{\sqrt{\sin ^4 x(\cos \alpha+\cot x \sin \alpha)}}$

$\frac{\operatorname{cosec}^2 x}{\sqrt{(\cos \alpha+\cot x \sin \alpha)}}$

$(\cos \alpha+\cot x \sin \alpha)=t$

Let $\Longrightarrow-\operatorname{cosec}^2 x \cdot \sin \alpha d x=d t$

$\therefore I=\frac{-1}{\sin \alpha} \int \frac{1}{\sqrt{t}} d t$

$=-\frac{1}{\sin \alpha} \int \mathrm{t}^{-\frac{1}{2}} \cdot \mathrm{dt}$

$=-\frac{2}{\sin \alpha}[\sqrt{\mathrm{t}}]+\mathrm{C}$

$=-\frac{2}{\sin \alpha}[\sqrt{(\cos \alpha+\cot \mathrm{x} \sin \alpha)}]+\mathrm{C}$

$=-\frac{2}{\sin \alpha}\left[\sqrt{\left(\cos \alpha+\frac{\cos x}{\sin x} \sin \alpha\right)}\right]+C$

$=-\frac{2}{\sin \alpha}\left[\sqrt{\frac{(\cos \alpha \sin x+\cos x \sin \alpha)}{\sin x}}\right]+C$

$\therefore I=\frac{-2}{\sin \alpha} \sqrt{\frac{\sin (x+\alpha)}{\sin x}}+C$

Posted by

HARSH KANKARIA

View full answer

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