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If I_{m, n}=\int \cos ^{m} x \cos n x d x then Find the value of I_{6,2}

Option: 1

I_{6, 2}= \frac{\cos ^{6} x \sin 2 x}{8}+\frac{4}{3} I_{5, 1}


Option: 2

I_{6, 2}= \frac{\cos ^{6} x \sin 2 x}{8}-\frac{4}{3} I_{5, 1}


Option: 3

I_{6, 2}= \frac{\cos ^{6} x \sin 2 x}{8}-\frac{3}{4} I_{5, 1}


Option: 4

I_{6, 2}= \frac{\cos ^{6} x \sin 2 x}{8}+\frac{3}{4} I_{5, 1}


Answers (1)

best_answer

 

 

Reduction Formula (Part 2) -

\mathbf{Integration\;of\;the\;type\;\;\;\int \cos^m x\; \cos nx\;dx}

-

\\\mathrm{Let\;I_{m,n}=\int\cos^m x\cos nx\;dx}\\\\\text{To\;evaluate\;this\;integral, we will use integration by parts method}\\ \mathrm{Here,\;take\;\cos^m x\;as\;first\;function\;and \;\cos nx\;as\;second\;function.}

\\\\\mathrm{\;\;\;\;\;\;\;\;\;}=\frac{\cos ^{m} x \sin n x}{n}-\frac{m}{n} \int \cos ^{m-1} x (-\sin x) \sin n x d x \\\mathrm{\;\;\;\;\;\;\;\;\;}=\frac{\cos ^{m} x \sin n x}{n}+\frac{m}{n} \int \cos ^{m-1} x\{\cos (n-1) x-\cos n x \cos x\} d x \\ \text { [using } \cos (n-1) x=\cos n x \cos x+\sin n x \sin x\\\mathrm{\;\;\;\;\;\;}\Rightarrow \sin x \sin n x=\cos (n-1) x-\cos n x \cos x] \\\\\mathrm{\;\;\;\;\;\;\;\;\;}= \frac{\cos ^{m} x \cos n x}{n}-\frac{m}{n} \int \cos ^{m} x \cos n x d x+\frac{m}{n} \int \cos ^{m-1} x \cos (n-1) x d x \\\mathrm{\;\;\;}I_{m, n}= \frac{\cos ^{m} x \cos n x}{n}-\frac{m}{n} I_{m, n}+\frac{m}{n} I_{m-1, n-1}

\\ {\Rightarrow \frac{m+n}{n} I_{m, n}= \frac{\cos ^{m} x \sin n x}{n}+\frac{m}{n} I_{m-1, n-1}} \\ {\text { or }\;\;\;\;\;\; \quad I_{m, n}= \frac{\cos ^{m} x \sin n x}{m+n}+\frac{m}{m+n} I_{m-1, n-1}}

Now put the value of m=6 and n=2

I_{6, 2}= \frac{\cos ^{6} x \sin 2 x}{8}+\frac{3}{4} I_{5, 1}

 

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