Question

interval cos^3 x sin x dx​

Answer 1

EXPLANATION.

∫Cos³x.Sinx dx.

As we know that,

By applying substitution method, we get.

Let, we assume that,

⇒ Cos(x) = t.

Differentiate w.r.t x, we get.

⇒ -Sin(x)dx = dt.

⇒ ∫Cos³x.Sinx.dx.

⇒ ∫t³(-dt).

⇒ -∫t³dt.

As we know that,

Formula of ⇒ ∫xⁿdx = xⁿ⁺¹/n + 1.

By applying this formula, we get.

⇒ -[t³⁺¹/3 + 1] + c.

⇒ -[t⁴/4] + c.

⇒ -t⁴/4 + c.

Substitute the value of t in equation, we get.

⇒ -Cos⁴(x)/4 + c.

                                                                                                                                     

MORE INFORMATION.

Some standard substitution.

(1) = √a² - x²  Or  1/√a² - x²

Substitute x = a sin∅.

(2) = √x² + a²  Or  1/√x² + a²

Substitute x = a tan∅  Or  x = a Sin(h)∅.

(3) = √x² - a²  Or  1/√x² - a²

Substitute x = a sec∅  Or  a = Cos(h)∅.

(4) = √x/a + x  Or  √a + x/x  Or  √x(x + a)  Or  1/√x(a + x).

Substitute x = a tan²∅.

(5) = √x/a - x  Or  √a - x/x  Or  1/√x(a - x).

Substitute x = a sin²∅.

(6) = √x/x - a  Or  √x - a/x  Or  √x(x - a)  Or  1/√x(x - a).

Substitute x = a sec²∅.

(7) = √a - x/a + x  Or  √a + x/a - x.

Substitute x = a Cos2∅.

(8) = √x - α/β - x  Or  √(x - α)(β - x),  (β > 0).

Substitute x = αcos²∅ + βsin²∅.

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\tt \ \: :  ⟼ \: Let \: I \: = \: \int \: ( {cos}^{3} x \: sinx )\: dx$

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$\tt \ \: :  ⟼ Now, \: put \: cosx \: = y$

$\tt \ \: :  ⟼ differentiate \: w.r.t. \: x \: we \: get$

$\tt \ \: :  ⟼ - \: sinx \: = \dfrac{dy}{dx}$

$\tt \ \: :  ⟼ \: sinx \: dx \: = - \: dy \:$

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$\tt \:  ⟼ So \: I = \int {y}^{3} ( - dy)$

$\tt\implies \:I = - \dfrac{ {y}^{4} }{4} + c \: \: \: ( \because \: \int {x}^{n} dx = \dfrac{ {x}^{ n+ 1} }{ n+ 1} + c)$

$\tt\implies \:I = - \dfrac{1}{4} {cos}^{4} x + c$

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