If -cos5 x dx-nC0-sin x-nC1-sin3x3-nC2-sin5x5-C- then the value of 2n-3C2 is equal towhere C is integration constant

∫cos^5 x dx=∫cos^4 x·cos x dx nbsp; =∫(1 sin^2 x)^2cos x dx =∫(1 t^2)^2 dt, where t=sin x =∫(1 2t^2+t^4)dt =t 2t^33+t^55+C =^2C_0·sin x ^2C_1·sin^3x3+^2C_2·s ...

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Standard X

Physics

Magnetic Field Due to Straight Current Carrying Conductor

If ∫cos5 x dx...

Question

If $\int {cos}^{5} x d x =^{n} C_{0} \cdot sin x -^{n} C_{1} \cdot \frac{{sin}^{3} x}{3} +^{n} C_{2} \cdot \frac{{sin}^{5} x}{5} + C,$ then the value of $^{(2 n + 3)} C_{2}$ is equal to
(where $C$ is integration constant)

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Solution

$\int {cos}^{5} x d x = \int {cos}^{4} x \cdot cos x d x$
$= \int (1 - {sin}^{2} x)^{2} cos x d x$
$= \int (1 - t^{2})^{2} d t,$ where $t = sin x$
$= \int (1 - 2 t^{2} + t^{4}) d t = t - \frac{2 t^{3}}{3} + \frac{t^{5}}{5} + C$
$=^{2} C_{0} \cdot sin x -^{2} C_{1} \cdot \frac{{sin}^{3} x}{3} +^{2} C_{2} \cdot \frac{{sin}^{5} x}{5} + C$
So, $n = 2$
$∴^{(2 n + 3)} C_{2} =^{7} C_{2} = 21$

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Magnetic Field Due to Straight Current Carrying Conductor

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If ∫cos5x dx=nC0⋅sinx−nC1⋅sin3x3+nC2⋅sin5x5+C, then the value of (2n+3)C2 is equal to (where C is integration constant)

∫cos5x dx=∫cos4x⋅cosx dx =∫(1−sin2x)2cosx dx =∫(1−t2)2 dt, where t=sinx =∫(1−2t2+t4)dt=t−2t33+t55+C =2C0⋅sinx−2C1⋅sin3x3+2C2⋅sin5x5+C So, n=2 ∴(2n+3)C2=7C2=21

If $\int {cos}^{5} x d x =^{n} C_{0} \cdot sin x -^{n} C_{1} \cdot \frac{{sin}^{3} x}{3} +^{n} C_{2} \cdot \frac{{sin}^{5} x}{5} + C,$ then the value of $^{(2 n + 3)} C_{2}$ is equal to
(where $C$ is integration constant)