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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\large\frac{1}{\sqrt{\sin^3x\sin(x+\alpha)}}$

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Toolbox:
  • If an integral function $f(x)=t,$ then $f'(x)dx=dt$ then $\int f(x)dx=\int tdt$
  • $\int x^n dx=\frac{x^{n+1}}{n+1}+c$
  • $\sin(A+B)=\sin A\cos B+\cos A\sin B$
Step 1:
Let $I=\int\large\frac{1}{\sqrt {\sin^3x\sin(x+\alpha)}}$
Let us consider $\sin^3x\sin(x+\alpha)$
$\sin^3x\sin(x+\alpha)=\sin^3x(\sin x\cos \alpha+\cos x\sin\alpha)$
$\qquad\qquad\;\;\;\;\quad=\sin^4x\cos\alpha+\sin^3x\cos x\sin\alpha$
$\qquad\qquad\;\;\;\;\quad=\sin^4x(\cos\alpha+\cot x\sin\alpha)$
Step 2:
Put $\cos\alpha+\cot x\sin\alpha=t$
On differentiating we get,
$-cosec^2x\sin\alpha\sqrt{t}dt=-\large\frac{1}{\sin\alpha}$$\int t^{\Large\frac{1}{2}}$
On integrating we get,
$-\large\frac{1}{\sin\alpha}[\large\frac{t^{\Large\frac{1}{2}}}{1/2}]$+c
$\Rightarrow-2cosec\alpha.\sqrt{t}+c$
Step 3:
Substituting for t we get,
$-2cosec\alpha(\cos \alpha+\cot x\sin\alpha)^{\Large\frac{1}{2}}$+c
$-2cosec\alpha(\cos \alpha+\large\frac{\cos x}{\sin x}\normalsize\sin\alpha)^{\Large\frac{1}{2}}$+c
$-2cosec\alpha(\large\frac{\sin x\cos \alpha+\cos x\sin\alpha}{\sin x})^{\Large\frac{1}{2}}$+c
$\sin(A+B)=\sin A\cos B+\cos A\sin B$
$\Rightarrow -2\sin\alpha.\sqrt{\large\frac{\sin(x+\alpha)}{\sin x}}$+c
answered Sep 10, 2013 by sreemathi.v
 
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