Subjective Type

Find the integral of $$\displaystyle \int \frac {\sec^2x}{co\sec^2x}dx$$

Solution

$$\displaystyle \int \frac {\sec^2x}{co\sec^2x}dx$$
$$\displaystyle=\int \frac {\frac {1}{\cos^2x}}{\frac {1}{\sin^2x}}dx$$
$$\displaystyle=\int \frac {\sin^2x}{\cos^2x}dx$$
$$\displaystyle=\int \tan^2x dx$$
$$\displaystyle=\int (\sec^2x-1)dx$$
$$\displaystyle=\int \sec^2xdx-\int 1 dx$$
$$=\tan x-x+C$$

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