We have an answer to "How to find the integral of ∫sin3xdx ?"

Anabella Gilbert

Answered question

2022-12-23

How to find the integral of

\int \sin^{3} x d x

Raymond Patel

Beginner2022-12-24Added 7 answers

\int \sin^{3} x d x = \int \sin x (1 - \cos^{2} x) d x (∵ \sin^{2} x = 1 - \cos^{2} x) \Rightarrow \int \sin^{3} x d x = \int \sin x d x - \int \sin x . \cos^{2} x d x

Solve the first integral.

\int \sin x d x = \cos x + C

Solve the second integral.
Let's suppose

\cos x = u

hence,

- \sin x d x = d u

replace the value,

- \int \sin x . \cos^{2} x d x = \int u^{2} d u \Rightarrow - \int \sin x . \cos^{2} x d x = \frac{u^{3}}{3} + C \Rightarrow - \int \sin x . \cos^{2} x d x = \frac{1}{3} \cos^{3} x + C

Combining the aforementioned,

\int \sin^{3} x d x = \cos x + \frac{1}{3} \cos^{3} x + C

Therefore the required value of

\int \sin^{3} x d x

\cos x + \frac{1}{3} \cos^{3} x + C

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