If Im-n-sinm xcosn x dx- then I5-4-

We know that, I_m,n=∫sin^m xcos^n x dx= sin^m 1xcos^n+1xm+n+m 1m+n· I_m 2,n Putting m=5, n=4 in the above equation, we get I_5,4= sin^4xcos^5x9+49· I_3,4 ...

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Standard X

Physics

Magnetic Field Due to Straight Current Carrying Conductor

If Im,n=∫sinm...

Question

If $I_{m, n} = \int {sin}^{m} x {cos}^{n} x d x,$ then $I_{5, 4} =$

- \frac{{sin}^{4} x {cos}^{5} x}{9} + \frac{4}{9} \cdot I_{4, 3}

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- \frac{{sin}^{4} x {cos}^{5} x}{9} + \frac{4}{9} \cdot I_{3, 4}

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- \frac{{sin}^{5} x {cos}^{4} x}{9} + \frac{4}{9} \cdot I_{3, 4}

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- \frac{{sin}^{5} x {cos}^{4} x}{9} + \frac{4}{9} \cdot I_{4, 3}

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Solution

The correct option is B $- \frac{{sin}^{4} x {cos}^{5} x}{9} + \frac{4}{9} \cdot I_{3, 4}$
We know that,
$I_{m, n} = \int {sin}^{m} x {cos}^{n} x d x = - \frac{{sin}^{m - 1} x {cos}^{n + 1} x}{m + n} + \frac{m - 1}{m + n} \cdot I_{m - 2, n}$
Putting $m = 5, n = 4$ in the above equation, we get
$I_{5, 4} = - \frac{{sin}^{4} x {cos}^{5} x}{9} + \frac{4}{9} \cdot I_{3, 4}$

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Magnetic Field Due to Straight Current Carrying Conductor

Standard X Physics

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If Im,n=∫sinmxcosnxdx, then I5,4=

The correct option is B −sin4xcos5x9+49⋅I3,4We know that, Im,n=∫sinmxcosnxdx=−sinm−1xcosn+1xm+n+m−1m+n⋅Im−2,n Putting m=5,n=4 in the above equation, we get I5,4=−sin4xcos5x9+49⋅I3,4

If $I_{m, n} = \int {sin}^{m} x {cos}^{n} x d x,$ then $I_{5, 4} =$