Question

What is the integral of `int (1-(tan x )^2)/ (sec (x)^2)`?

Answer 1

`1/2 (2x+sin (2x)) - x + C`

Explanation:

Using the following relation,
`-\tan ^2x\=1-\sec ^2x`

`int (1+1-sec^2(x))/sec^2(x)dx`

Applying sum rule,
`int f(x)+- g(x) dx=int f(x)dx +- intg(x) dx`

`\int \frac{1}{sec ^2x)dx+\int \frac{1}{\sec ^2(x\)}dx-\int \frac{\sec ^2(x\)}{\sec ^2(x\)}dx`....... eq(i)

`\int \frac{1}{\sec ^2(x\)}dx=\frac{1}{4}(2x+\sin (2x))`

`\int \frac{1}{\sec ^2(x\)}dx=\frac{1}{4}(2x+\sin (2x))`

`\int \frac{\sec ^2(x\)}{\sec ^2\(x)}dx=x`

Substituting the values in eqn (i) we get,
`\frac{1}{4}(2x+\sin \(2x)+\frac{1}{4}(2x+sin (2x))-x`

Simplifying we get
`1/2 (2x+sin (2x)) - x + C`