Question

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Answer 1

Question :–

$\\\bf \to Integrate :\int ({\sin}^{3}x) ({ \cos}^{2} x).dx$

ANSWER :–

GIVEN :–

• A function –

$\\\bf \implies I = \int ({\sin}^{3}x) ({ \cos}^{2} x).dx$

TO FIND :–

• I = ?

SOLUTION :–

$\\\bf \implies I = \int ({\sin}^{3}x) ({ \cos}^{2} x).dx$

$\\\bf \implies I = \int ( \sin x)({\sin}^{2}x) ({ \cos}^{2} x).dx$

• Using identity –

$\\\bf \bigstar \:\:{\sin^{2} }(x) + { \cos} ^{2} (x) = 1$

• So that –

$\\\bf \implies I = \int ( \sin x)(1 - {\cos}^{2}x) ({ \cos}^{2} x).dx$

• Now put cos x = t –

$\\\bf \implies \cos(x) = t$

• Differentiate with respect to 't' –

$\\\bf \implies - \sin(x) \frac{dx}{dt} =1$

$\\\bf \implies - \sin(x) dx=dt$

$\\\bf \implies \sin(x) dx=-dt$

• Now –

$\\\bf \implies I = - \int(1 - {t}^{2}) ({t}^{2} ).dt$

$\\\bf \implies I = \int({t}^{2} - 1) ({t}^{2} ).dt$

$\\\bf \implies I = \int({t}^{4} - {t}^{2} ).dt$

• Using identity –

$\\\bf \implies \int{x}^{n}.dx = \dfrac{ {x}^{n + 1} }{n + 1}$

$\\\bf \implies I = \dfrac{ {t}^{5} }{5} - \dfrac{ {t}^{3} }{3} + c$

• Now replace 't' –

$\\\bf \implies I = \dfrac{{ \cos}^{5}(x) }{5} - \dfrac{ { \cos}^{3}(x) }{3} + c$

Answer 2

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