# Find the integral:{eq}\int_{0}^{1} \tan^{2}x \sec^{2} x dx {/eq}

## Question:

Find the integral:{eq}\int_{0}^{1} \tan^{2}x \sec^{2} x dx {/eq}

## Definite Integral :

The definite integral of the type {eq}\int_{a}^{b}f(x)f'(x)\ dx {/eq} can be evaluated by using the substitution method.

In this method we take {eq}f(x)=t {/eq} (say) so that we get {eq}f'(x)\ dx=dt. {/eq}

Thus, the integral becomes easy to solve. The limits of the integral would be changed accordingly.

Let the given integral be {eq}I {/eq} therefore, we get:

{eq}I = \int_{0}^{1} \tan^{2}x \sec^{2} x dx {/eq}

Take {eq}\tan x = t {/eq} so that {eq}\sec^{2} x dx = dt. {/eq}

At {eq}x=0 {/eq} we get {eq}t=0 {/eq} and at {eq}x=1 {/eq} we get {eq}t=\tan 1 {/eq}

Therefore, the given integral can be written as:

{eq}\\\\\begin{align*} I & = \int_{0}^{\tan 1}t^{2}\ dt \\\\& = \left [ \frac{t^{3}}{3} \right ]_{0}^{\tan 1} \\\\& = \left [ \frac{\tan^{3}1}{3}-0 \right ] \ \ \ \ \ \texttt{(substituting the limits)} \\\\& = \frac{\tan^{3}1}{3} \\\\& = \frac{3.777}{3} \\\\& \approx 1.26 \\\\\end{align*} {/eq} 