Let In-sinnx dx and Jm-cosmx dx- If I4-J4-Ax-Bsin4x-C- then AB is equal towhere C is integration constant

I_4= sin^3xcos x4+34∫sin^2x dx J_4=cos^3xsin x4+34∫cos^2x dx I_4+J_4=14sin x·cos x(cos^2x sin^2x)+34∫ dx =34x+116sin4x+C So, A=34, B=116 Hence, AB=12 ...

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Byju's Answer

Standard X

Physics

Magnetic Field Due to Straight Current Carrying Conductor

Let In=∫sinnx...

Question

Let $I_{n} = \int {sin}^{n} x d x$ and $J_{m} = \int {cos}^{m} x d x .$ If $I_{4} + J_{4} = A x + B sin 4 x + C,$ then $\frac{A}{B}$ is equal to
(where $C$ is integration constant)

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Solution

$I_{4} = - \frac{{sin}^{3} x cos x}{4} + \frac{3}{4} \int {sin}^{2} x d x$
$J_{4} = \frac{{cos}^{3} x sin x}{4} + \frac{3}{4} \int {cos}^{2} x d x$
$I_{4} + J_{4} = \frac{1}{4} sin x \cdot cos x ({cos}^{2} x - {sin}^{2} x) + \frac{3}{4} \int d x$
$= \frac{3}{4} x + \frac{1}{16} sin 4 x + C$
So, $A = \frac{3}{4}, B = \frac{1}{16}$
Hence, $\frac{A}{B} = 12$

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Let In=∫sinnx dx and Jm=∫cosmx dx. If I4+J4=Ax+Bsin4x+C, then AB is equal to (where C is integration constant)

I4=−sin3xcosx4+34∫sin2x dx J4=cos3xsinx4+34∫cos2x dx I4+J4=14sinx⋅cosx(cos2x−sin2x)+34∫dx =34x+116sin4x+C So, A=34,B=116 Hence, AB=12

Let $I_{n} = \int {sin}^{n} x d x$ and $J_{m} = \int {cos}^{m} x d x .$ If $I_{4} + J_{4} = A x + B sin 4 x + C,$ then $\frac{A}{B}$ is equal to
(where $C$ is integration constant)