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

1 Answer
Feb 14, 2016

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