We have an answer to "Evaluate the following integrals. ∫sec−2xtan3xdx"

Kye

Answered question

2021-10-12

Evaluate the following integrals.

\int \sec^{- 2} x \tan^{3} x d x

cheekabooy

Skilled2021-10-13Added 83 answers

The given integral is

I = \int \sec^{- 2} x \tan^{3} x d x

We have:

I = \int \sec^{- 2} x \tan^{3} x d x

I = \int \frac{\tan^{3} x}{\sec^{2} x} d x

I = \int \frac{\sin^{3} x \cos^{2} x}{\cos^{3} x} d x

I = \int \frac{\sin^{3} x}{\cos x} d x

I = \int \frac{\sin x \sin^{2} x}{\cos x} d x

let

\cos x = t \cos^{2} x = t^{2} \Rightarrow 1 - \sin^{2} x = t^{2}; \sin^{2} x = 1 - t^{2}

- \sin x d x = d t

\Rightarrow I = - \int \frac{(1 - t^{2})}{t} d t

\Rightarrow I = - \int (\frac{1}{t} - \frac{t^{2}}{t}) d t

\Rightarrow I = \int (t - \frac{1}{t}) d t

\Rightarrow I = \int t d t - \int \frac{1}{t} d t

( using

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

and

\int \frac{1}{x} d x = \ln | x | + C

)

\Rightarrow I = \frac{t^{2}}{2} - \ln | (t) | + C

\Rightarrow I = \frac{\cos^{2} x}{2} - \ln | \cos x | + C

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