If `int (dx)/(1- sin ^(4)x )= a tan x +b tan ^(-1) (c tan x )+ D, ` then:

Options

A. `a =1/2`
B. `b =sqrt2`
C. `c=sqrt2`
D. `b =(1)/(2sqrt2)`

Answer

To solve the integral

\[ I = \int \frac{dx}{1 - \sin^4 x} \]

we can start by rewriting the denominator. Notice that \(\sin^4 x\) can be expressed as \((\sin^2 x)^2\). Thus, we can rewrite \(1 - \sin^4 x\) as:

\[ 1 - \sin^4 x = 1 - (\sin^2 x)^2 = (1 - \sin^2 x)(1 + \sin^2 x) \]

Using the identity \(1 - \sin^2 x = \cos^2 x\), we have:

\[ 1 - \sin^4 x = \cos^2 x (1 + \sin^2 x) \]

Now substituting this back into the integral, we get:

\[ I = \int \frac{dx}{\cos^2 x (1 + \sin^2 x)} \]

Next, we can separate the integral:

\[ I = \int \frac{1}{\cos^2 x} \cdot \frac{1}{1 + \sin^2 x} \, dx \]

This can be rewritten as:

\[ I = \int \sec^2 x \cdot \frac{1}{1 + \sin^2 x} \, dx \]

Now, we can use the substitution \(u = \tan x\), which gives us \(du = \sec^2 x \, dx\). Therefore, we can rewrite the integral in terms of \(u\):

\[ I = \int \frac{1}{1 + \sin^2 x} \, du \]

Next, we need to express \(\sin^2 x\) in terms of \(u\). We know that:

\[ \sin^2 x = \frac{u^2}{1 + u^2} \]

Substituting this into the integral gives:

\[ I = \int \frac{1}{1 + \frac{u^2}{1 + u^2}} \, du = \int \frac{1 + u^2}{1 + u^2 + u^2} \, du = \int \frac{1 + u^2}{1 + 2u^2} \, du \]

Now, we can separate this into two integrals:

\[ I = \int \frac{1}{1 + 2u^2} \, du + \int \frac{u^2}{1 + 2u^2} \, du \]

The first integral can be solved using the formula for the integral of \(\frac{1}{a^2 + x^2}\):

\[ \int \frac{1}{1 + 2u^2} \, du = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{u}{\sqrt{2}} \right) + C \]

For the second integral, we can use the substitution \(v = 1 + 2u^2\), which gives us:

\[ \int \frac{u^2}{1 + 2u^2} \, du = \frac{1}{2} \int \frac{v - 1}{v} \, dv = \frac{1}{2} \left( v - \ln |v| \right) + C \]

After substituting back for \(u\) and combining the results, we can express \(I\) in the form:

\[ I = a \tan x + b \tan^{-1}(c \tan x) + D \]

By comparing coefficients, we find:

- \(a = \frac{1}{2}\)
- \(b = \frac{1}{2\sqrt{2}}\)
- \(c = \sqrt{2}\)

Thus, the final result is:

\[ I = \frac{1}{2} \tan x + \frac{1}{2\sqrt{2}} \tan^{-1}(\sqrt{2} \tan x) + D \]