Question

Hi guys!


ಇಲ್ಲ ಯಾರಾದ್ರೂ ಕನ್ನಡಿಗರು ಇದ್ದೀರಾ!?​

Answer 1

Explanation:

ಹೌದು ನನಗೆ ಕನ್ನಡ ಗೊತ್ತಿಲ್ಲ

yes we are there ❤️

Answer 2

• Your welcome sis/bro(◍•ᴗ•◍)✧*。
• Myself Dua
• What's your name
• Nice to meet you
• Can we become friends

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

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.

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

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

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