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# Find the integrals of the functions$\frac{\cos2x-\cos2\alpha}{\cos x-\cos\alpha}$

Can you answer this question?

Toolbox:
• $(i)\cos C-\cos D=-2\sin\frac{(C+D)}{2}\sin\frac{(C-D)}{2}.$
• $(ii)\sin x=2\sin\frac{x}{2}\cos \frac{x}{2}.$
Given $I=\int\frac{\cos2x-\cos2\alpha}{\cos x-\cos \alpha}dx.$

This can be written as,

$I=\int\frac{-2\sin\frac{2x+2\alpha}{2}\sin\frac{2x-2\alpha}{2}}{-2\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}dx.$

$\;\;\;=\int\frac{\sin(x+\alpha)\sin(x-\alpha)}{\sin\frac{(x+\alpha)}{2}.\sin\frac{(x-\alpha)}{2}}dx.$

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

$\;\;\;=\frac{\int\begin{bmatrix}\sin(\frac{x+\alpha}{2})\cos(\frac{x+\alpha}{2})\end{bmatrix}\begin{bmatrix}\frac\sin(\frac{x-\alpha}{2})\cos(\frac{x-\alpha}{2})\end{bmatrix}}{\sin\big(\frac{x+\alpha}{2}\big).\sin\big(\frac{x-alpha}{2}\big)}dx.$

Cancelling the common terms we get,

$\\;\;=\int4\cos\bigg(\frac{x+\alpha}{2}\bigg).\cos\bigg(\frac{x-\alpha}{2}\bigg)dx.$

But this is of the form $2\cos C .\cos D=\cos (C+D)+\cos(C-D).$

$\;\;\;=2\int[\cos\big(\frac{x+\alpha}{2}+\frac{x-\alpha}{2}\big)+\cos\big(\frac{x+\alpha}{2}-\frac{x-\alpha}{2}\big)]dx.$

$\;\;\;=2\int\cos \alpha dx+\int\cos \alpha dx.$

On integrating we get,

$\int\frac{\cos2x-\cos2\alpha}{\cos x-\cos \alpha}dx=2\sin x+2x\cos\alpha+c.$

$\;\;\;\;\;\;\qquad\qquad\;\;\;=2[\sin x+x\cos \alpha]+c$

answered Jan 31, 2013