Question
Evaluate the following integrals:
$\int\limits^{{\pi}}_0\cos^5\text{x dx}$
$\int\limits^{{\pi}}_0\cos^5\text{x dx}$
✓
Answer
Let $\text{I}=\int\limits^{{\pi}}_0\cos^5\text{x dx}$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(\cos^2\text{x}\big)^2\text{dx}$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(1-\sin^2\text{x}\big)^2\text{dx}$
Let $\sin\text{x}=\text{t},$ then $\cos\text{x dx}=\text{dt}$
When, $\text{x}\rightarrow0;\text{ t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{ t}\rightarrow0$
Therefore,
$\text{I}=\int\limits^0_0\big(1-\text{t}^2\big)^2\text{dt}$
$\text{I}=0$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(\cos^2\text{x}\big)^2\text{dx}$
$=\int\limits^{{\pi}}_0\cos\text{x}\big(1-\sin^2\text{x}\big)^2\text{dx}$
Let $\sin\text{x}=\text{t},$ then $\cos\text{x dx}=\text{dt}$
When, $\text{x}\rightarrow0;\text{ t}\rightarrow0$ and $\text{x}\rightarrow\pi;\text{ t}\rightarrow0$
Therefore,
$\text{I}=\int\limits^0_0\big(1-\text{t}^2\big)^2\text{dt}$
$\text{I}=0$
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