Question

If $\int {\sin }^{4}xdx=-\frac{\cos x{\left(\sin x\right)}^3}{a}+\frac{b}{a}(-\frac{\cos x\left(\sin x\right)}{2}+\frac{1}{2}x)+c$, then $a+b$ is equal to

• A. 7
• B. 4
• C. -6
• D. 5

Answer 1

The correct option is A 7

Let us consider $I_n=\int {\sin }^{n}xdx=\int \sin x{\sin }^{n-1}xdx$

We know that, $\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

$I_n=-\cos x{\left(\sin x\right)}^{n-1}+(n-1)\int {\left(\sin x\right)}^{n-2}{\cos }^{2}xdx$

$=-\cos x{\left(\sin x\right)}^{n-1}+(n-1)\int {\left(\sin x\right)}^{n-2}(1-{\sin }^{2}x)dx$

$\Rightarrow I_n=-\cos x{\left(\sin x\right)}^{n-1}+(n-1)I_{n-2}-(n-1)I_n$

$\Rightarrow I_n=-\frac{\cos x{\left(\sin x\right)}^{n-1}}{n}+\frac{n-1}{n}{I}_{n-2}(n\ge 2)$

Hence $I_4=-\frac{\cos x{\left(\sin x\right)}^3}{4}+\frac{3}{4}(-\frac{\cos x\left(\sin x\right)}{2}+\frac{1}{2}x)+c$

Comparing with $\int {\sin }^{4}xdx=-\frac{\cos x{\left(\sin x\right)}^3}{a}+\frac{b}{a}(-\frac{\cos x\left(\sin x\right)}{2}+\frac{1}{2}x)+c$

We get, $b=3anda=4$

Then $a+b=7$