# 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.

## Answer and Explanation: 1

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}

#### Learn more about this topic:

Evaluating Definite Integrals Using the Fundamental Theorem

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Chapter 16 / Lesson 2
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In calculus, the fundamental theorem is an essential tool that helps explain the relationship between integration and differentiation. Learn about evaluating definite integrals using the fundamental theorem, and work examples to gain understanding.