AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

∫

\sin^5

\cos^5

x dx =

$\frac{\cos^{6} x}{60}$ (6 $\sin^{4}$ x + 3 $\sin^{2}$ x + 1) + c
- $\frac{\sin^{6} x}{60}$ (6 $\cos^{4}$ x + 3 $\cos^{2}$ x + 1) + c
- $\frac{\cos^{6} x}{60}$ (6 $\sin^{4}$ x + 3 $\sin^{2}$ x + 1) + c
$\frac{\sin^{6} x}{60}$ (6 $\cos^{4}$ x + 3 $\cos^{2}$ x + 1) + c
c and d

Solution:

Consider the integral.

I \;\; =\int\sin^{5} x \cos^{5} x d x

\;\; =\int\sin^{4} x \sin x \cos^{5} x d x

\;\; =\int(1-\cos^{2} x)^{2} \cos^{5} x d x

Let

\cos x \;\; =t

-\sin xdx \;\; =d t

Therefore

I \;\; =\int(1-t)^{2} t^{5}(-d t)

\;\; =-\left[\;\frac{t^{10}}{10}-2\;\frac{t^{8}}{8}+\;\frac{t^{6}}{6}\right]+c

\;\; =-\;\frac{t^{6}}{60}[6 t^{4}-15 t^{2}+10]+c

\;\; =-\;\frac{\cos^{6} x}{60}[6 \cos^{4} x-15 \cos^{2} x+10]+c

Solve further

I \;\; =-\;\frac{\cos^{6} x}{60}[6 \sin^{4} x-12 \sin^{2} x+6-15+15 \sin^{2} x+10]+c

\;\; =-\;\frac{\cos^{6} x}{60}[6 \sin^{4} x-3 \sin^{2} x+1]+c

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