Question

Integration of cos^99x upper limit 2π lower limit 0

Answer 1

Answer:

your question answer is 0 bro

because integration of cosx is sinx

in limit 0 to 2pai is sinx is also zero

Answer 2

Answer:

$\int_{0}^{2 \pi} \cos ^{99} \mathrm{xdx}=0$

Step-by-step explanation:

$\int_{0}^{2 \pi} \cos ^{99} \mathrm{xdx}$

Using reduction formula

$\int \cos ^{\mathrm{m}} x d x=\sin x \frac{\cos ^{\mathrm{m}-1} \mathrm{x}}{\mathrm{m}}+\frac{\mathrm{m}-1}{\mathrm{~m}} \int \cos ^{\mathrm{m}-2} \mathrm{xdx} \\ \int \cos ^{99} x d x=\sin x \frac{\cos ^{98} x}{99}+\frac{98}{99} \int \cos ^{97} x d x$

$\int_{0}^{2 \pi} \cos ^{99} x d x=\left[\sin x \frac{\cos ^{98} x}{99}\right]_{0}^{2 \pi}+\frac{98}{99} \int_{0}^{2 \pi} \cos ^{97} x d x$

$\int_{0}^{2 \pi} \cos ^{99} x d x=\frac{98}{99} \int_{0}^{2 \pi} \cos ^{97} x d x$

Again applying reduction formula,

$\int_{0}^{2 \pi} \cos ^{97} x d x=\left[\sin x \frac{\cos ^{96} x}{97}\right]_{0}^{2 \pi}+\frac{96}{99} \int_{0}^{2 \pi} \cos ^{95} x d x \\ \int_{0}^{2 \pi} \cos ^{97} x d x=\frac{96}{99} \int_{0}^{2 \pi} \cos ^{95} x d x$

$\therefore \int_{0}^{2 \pi} \cos ^{99} x d x=\frac{98}{99} \frac{96}{99} \int_{0}^{2 \pi} \cos ^{95} x d x$

$\therefore \int_{0}^{2 \pi} \cos ^{99} \mathrm{xdx}=\frac{98}{99} \frac{96}{99} 99 \cdot \cdots \cdot \frac{2}{99} \int_{0}^{2 \pi} \cos \mathrm{xdx}$

$\Rightarrow \int_{0}^{2 \pi} \cos ^{99} \mathrm{xdx}=\frac{98}{99} \frac{96}{99} 9 \frac{94}{99} \cdots \frac{2}{99}[\sin \mathrm{x}]_{0}^{2 \pi} \mathrm{dx} \\ \Rightarrow \int_{0}^{2 \pi} \cos ^{99} \mathrm{xdx}=0$