Question

Integrate with respect to $x$
i) ${\sin }^{2}x$
ii) ${\cos }^{3}x$
iii) $4\sin x\cos \frac{x}{2}\cos \frac{3x}{2}$
iv) $\frac{1}{\sqrt{x+3}-\sqrt{x+2}}$

Answer 1

$\left(i\right)\int {\sin }^{2}xdx$

$=\frac{1}{2}\int 2{\sin }^{2}xdx$

We know that $2{\sin }^{2}x=1-\cos 2x$

$=\frac{1}{2}\int (1-\cos 2x)dx$

$=\frac{1}{2}[x-\frac{\sin 2x}{2}]+c$

$=\frac{x}{2}-\frac{\sin 2x}{4}+c$

$\left(ii\right)\int {\cos }^{3}xdx$

$=\frac{1}{4}\int 4{\cos }^{3}xdx$

We know that $4{\cos }^{3}x=\cos 3x+3\cos x$

$=\frac{1}{4}\int (\cos 3x+3\cos x)dx$

$=\frac{1}{4}[\frac{\sin 3x}{3}+3\sin x]+c$

$=\frac{\sin 3x}{12}+\frac{3}{4}\sin x+c$

$\left(iii\right)\int 4\sin x\cos \frac{x}{2}\cos \frac{3x}{2}dx$

$=\int 2\sin x\left(2\cos \frac{x}{2}\cos \frac{3x}{2}\right)dx$

We know that $2\cos A\cos B=\cos (A+B)+\cos (A-B)$

$=\int 2\sin x(\cos (\frac{x}{2}+\frac{3x}{2})+\cos (\frac{x}{2}-\frac{3x}{2}))dx$

$=\int 2\sin x(\cos \left(\frac{4x}{2}\right)+\cos \left(\frac{-2x}{2}\right))dx$

$=\int (2\sin x\cos 2x+2\sin x\cos x)dx$

We know that $2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$=\int (\sin (x+2x)+\sin (x-2x)+\sin 2x)dx$

$=\int (\sin 3x-\sin x+\sin 2x)dx$

$=-\frac{\cos 3x}{3}+\cos x-\frac{\cos 2x}{2}+c$

$\left(iv\right)\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}}dx$

$=\int \frac{1}{\sqrt{x+3}-\sqrt{x+2}}\times \frac{\sqrt{x+3}+\sqrt{x+2}}{\sqrt{x+3}+\sqrt{x+2}}dx$

$=\int \frac{\sqrt{x+3}+\sqrt{x+2}}{x+3-x-2}dx$

$=\int (\sqrt{x+3}+\sqrt{x+2})dx$

$=\frac{{(x+3)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{{(x+2)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c$

$=\frac{{(x+3)}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{{(x+2)}^{\frac{3}{2}}}{\frac{3}{2}}+c$

$=\frac{2}{3}{(x+3)}^{\frac{3}{2}}+\frac{2}{3}{(x+2)}^{\frac{3}{2}}+c$