Question

integral sin²x*sinx​

Answer 1

Answer:

$\sf I = \dfrac{cos^{3}\:x }{3}-cos\:x+C$

Step-by-step explanation:

Given:

• sin² x × sin x

To Find:

$\displaystyle \sf \int\limits {sin^{2} x\: . \: sin\:x} \, dx$

Solution:

$\displaystyle \sf \int\limits {sin^{2} x\: . \: sin\:x} \, dx$

We know that,

sin² x = 1 - cos² x

Substituting it,

$\displaystyle \sf \int\limits {(1-cos^{2} x)\: . \: sin\:x} \, dx$

Now let us take,

t = cos x

Differentiating on both sides with respect to x,

dt = -sin x dx

-dt = sin x dx

Substituting in the above equation we get,

$\displaystyle \sf \int\limits {(1-t^{2} )} \, .-dt$

$\implies \displaystyle \sf \int\limits {(t^{2} -1)} \, .dt$

Now we know that,

$\displaystyle \sf \int\limits {x^n} \, dx =\dfrac{x^{n+1}}{n+1}$

$\displaystyle \sf \int\limits {1} \, dx =x+C$

Applying the identities,

$\implies \sf \dfrac{t^3}{3} -t + C$

Give back the value of t,

$\implies \boxed{ \sf \dfrac{cos^{3}\:x }{3}-cos\:x+C}$