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What is the reduction formula for

\int \sin ^{m}x\cos ^{n}xdx\, ;\, if\; k=\frac{-\sin ^{m-1}x\cos ^{n+1}x}{n+1}

  • Option 1)

    I\left ( m,n \right )=k+\frac{m+1}{n-1}I_{\left ( m-2,n+2 \right )}

  • Option 2)

    I\left ( m,n \right )=k+\frac{m-1}{n+1}I_{\left ( m-2,n+2 \right )}

     

  • Option 3)

    I\left ( m,n \right )=k+\frac{m}{n}I_{\left ( m-2,n+2 \right )}

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we learned,

 

Reduction formulae -

\int sin^{n}xdx=I_{n}   (let)

Then nI_{n}-(n-1)I_{n-2}=- \sin^{n-1}x\cos x

 

-

 

 I_{\left ( m,n \right )}=\int \sin ^{m}x\cos ^{n}xdx

I_{\left ( m,n \right )}=\int \sin ^{m-1}x \sin x\cos ^{n}xdx

=\sin ^{m-1}x\int \sin x\cos ^{n}x-\left ( m-1 \right )\int \sin ^{m-2}x\cos ^{n}xdx

=\sin ^{m-1}x\frac{\left [ -\cos ^{n+1}x \right ]}{n+1}-\left ( m-1 \right )\int \sin ^{m-2}x\cos x\left ( \frac{-\cos ^{m+1}}{m+1} \right )

=\frac{-\sin ^{m-1}x\cos ^{n+1}x}{n+1}+\frac{m-1}{n+1}I_{m-2,n+2}


Option 1)

I\left ( m,n \right )=k+\frac{m+1}{n-1}I_{\left ( m-2,n+2 \right )}

Option 2)

I\left ( m,n \right )=k+\frac{m-1}{n+1}I_{\left ( m-2,n+2 \right )}

 

Option 3)

I\left ( m,n \right )=k+\frac{m}{n}I_{\left ( m-2,n+2 \right )}

Option 4)

none of these

Posted by

gaurav

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