# Find the integral. \int sec^4 (4x) dx

## Question:

Find the integral.

{eq}\displaystyle \int \sec^4 (4x)\,dx {/eq}

## Substitution Method Of Solving Integrals:

There is a set of formulas that are used to solve integrals with different types of integrands. If the given integrand doesn't suit any of them, then we assume a part of the integrand to be u and we find du. Then we substitute the values in the given integral and solve it. This method is called the "substitution method" or "reverse way of chain rule of derivatives".

## Answer and Explanation:

The given integral is:

$$\int \sec^4 (4x)\,dx $$

We solve this using the substitution method:

$$\text{Let } 4x=u \\ \text{Then } 4 \, dx = du \\ dx = \dfrac{1}{4}du $$

Substitute this in the given integral:

$$\begin{align} \int \sec^4 (u)\, \left( \dfrac{1}{4} du \right) & = \dfrac{1}{4} \int \sec^4u du \\ &= \dfrac{1}{4} \int (\sec^2u) (\sec^2u) du \\ &= \dfrac{1}{4} \int (\sec^2u) (1+tan^2u)du & (\because \sec^2u - \tan^2u=1) \\ &= \dfrac{1}{4} \int (\sec^2u + \sec^2u \tan^2u) du \\ &= \dfrac{1}{4} \int \sec^2u du + \dfrac{1}{4} \int \sec^2u \tan^2u du \\ &= \dfrac{1}{4} (\tan u) + \dfrac{1}{4} \int \sec^2u \tan^2u du & \rightarrow (1) \end{align} $$

Now we will solve the integral in (1) using the substitution method again:

$$\text{Let } \tan u = v \\ \sec^2u du = dv $$

Substitute these value in the integral of (1):

$$\begin{align} \dfrac{1}{4} (\tan u) + \dfrac{1}{4} \int \sec^2u \tan^2u du & = \dfrac{1}{4} (\tan u) + \dfrac{1}{4} \int v^2 dv \\ &= \dfrac{1}{4} (\tan u) + \dfrac{1}{4} \left( \dfrac{v^3}{3}\right) +C \\ &= \dfrac{1}{4} (\tan u) + \dfrac{1}{12} v^3+C \\ & = \boxed{\mathbf{ \dfrac{1}{4} (\tan (4x)) + \dfrac{1}{12} (\tan^3 (4x))}} +C & \text{(Substituted the values of u and v back)} \end{align} $$

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