Math, asked by aasiyaanjum2000, 2 months ago

the coefficient of highest power of x or y is constant then curve has * ​

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Answered by amansharma264
8

EXPLANATION.

\implies \displaystyle \int sin^{3} x \ cos^{2} x \ dx

As we know that,

Formula of :

⇒ sin²x + cos²x = 1.

⇒ cos²x = 1 - sin²x

Using this formula in equation, we get.

\implies \displaystyle \int sin^{2} x \ cos^{2} x \ sinx \ dx

\implies \displaystyle \int (1 - cos^{2}x) \ cos^{2}x \ sin x \ dx

By using the substitution method, we get.

Let we assume that,

⇒ cos(x) = t.

⇒ - sin(x)dx = dt.

⇒ sin(x)dx = -dt.

Put the value in this equation, we get.

\implies \displaystyle \int (1 - t^{2} )(t^{2} )(-dt)

\implies \displaystyle \int (t^{2} - t^{4} )(-dt)

\implies \displaystyle \int (t^{4} - t^{2} )(dt)

\implies \displaystyle \int t^{4} dt \ - \int t^{2} dt

\implies \displaystyle \dfrac{t^{5} }{5} \ - \dfrac{t^{3} }{3} \ + C

Put the value of t = cos(x) in the equation, we get.

\implies \displaystyle \dfrac{cos^{5}x }{5} \ - \dfrac{cos^{3} x}{3} \ + C

\implies \displaystyle \dfrac{-1}{3} cos^{3} x \ + \dfrac{1}{5} cos^{5} x \ + C

Option [B] is correct answer.

                                                                                                                     

MORE INFORMATION.

Basic property of indefinite integration.

\implies \displaystyle \int k \ f(x)dx = k \int f(x)dx

\implies \displaystyle \int \bigg[ f_{1}(x) \pm f_{2}(x) \pm f_{3}(x) \pm . . . . \pm f_{n}(x) \bigg]dx \ = \int f_{1}(x)dx \pm \int f_{2}(x)dx \pm \int f_{3}(x)dx \pm . . . \pm \int f_{n}(x)dx

Integration of function f(ax + b).

\implies \displaystyle \int f(x)dx = f(x) + C.

\implies \displaystyle \int f(ax + b)dx = \dfrac{f(ax + b)}{a} \ + C

Proof.

\implies \displaystyle \int f(ax + b) dx

By using the substitution method, we get.

Let we assume that,

⇒ ax + b = t.

⇒ a dx = dt.

Put the value in the equation, we get.

\implies \displaystyle \int f(t) \dfrac{dt}{a}

\implies \displaystyle \dfrac{1}{a} \int f(t)dt

\implies \displaystyle \dfrac{1}{a} f(ax + b) + C \ = \dfrac{f(ax + b)}{a}  + C

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