Math, asked by sukhada6395, 8 months ago

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

Answers

Answered by Anonymous
2

Answer:

your question answer is 0 bro

because integration of cosx is sinx

in limit 0 to 2pai is sinx is also zero

Answered by ravilaccs
0

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$$

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