Question

Evaluate: $\int \dfrac{1}{\sin^{4} x+\sin^{2} x\cos^{2} x+\cos^{4} x} dx.$

Answer 1

Hint: In this question it is given that that we have to evaluate

 $$\int \dfrac{1}{\sin^{4} x+\sin^{2} x\cos^{2} x+\cos^{4} x} dx.$$

So to find the solution we need to use some formulas, which are

$$\left( a+b\right)^{2} =a^{2}+2ab+b^{2}$$.......(1)

$$\sin^{2} \theta +\cos^{2} \theta =1$$............(2)

After that we have to use the substitution method in order to get the solution.

Complete step-by-step answer:

Let,

$$I=\int \dfrac{1}{\sin^{4} x+\sin^{2} x\cos^{2} x+\cos^{4} x} dx$$

Adding and Subtracting $\sin^{2} x\cos^{2} x$ in Denominator, we get

$$=\int \dfrac{1}{\sin^{4} x+2\sin^{2} x\cos^{2} x+\cos^{4} x-\sin^{2} x\cos^{2} x} dx$$

$$=\int \dfrac{1}{\left( \sin^{2} x\right)^{2} +2\sin^{2} x\cos^{2} x+\left( \cos^{2} x\right)^{2} -\sin^{2} x\cos^{2} x} dx$$

since, Now by using formula (1) where $$a=\sin^{2} x$$ and $$b=\cos^{2} x$$, Now this is $$\left( a+b\right)^{2}$$ formula, we get,

$$I=\int \dfrac{1}{\left( \sin^{2} x+\cos^{2} x\right)^{2} -\sin^{2} x\cos^{2} x} dx$$

$$=\int \dfrac{1}{1^{2}-\sin^{2} x\cos^{2} x} dx$$ using formula (2)

$$=\int \dfrac{1}{1-\sin^{2} x\cos^{2} x} dx$$

$$=\int \dfrac{4}{4-4\sin^{2} x\cos^{2} x} dx$$ multiplying numerator and denominator by 4

$$=\int \dfrac{4}{4-2^{2}\sin^{2} x\cos^{2} x} dx$$

$$=\int \dfrac{4}{4-\left( 2\sin x\cos x\right)^{2} } dx$$

we know that, $$ 2\sin x\cos x\ = \sin 2x$$, By substituting this, we get

$$=\int \dfrac{4}{4-\left( \sin 2x\right)^{2} } dx$$

$$=\int \dfrac{4}{4-\sin^{2} 2x} dx$$

$$=\int \dfrac{4}{3+1-\sin^{2} 2x} dx$$

$$=\int \dfrac{4}{3+\cos^{2} 2x} dx$$ since, $$1-\sin^{2} \theta =\cos^{2} \theta$$

Now multiplying numerator and denominator by $$\sec^{2} 2x$$, we get,

$$I=\int \dfrac{4\sec^{2} 2x}{3\sec^{2} 2x+\sec^{2} 2x\cos^{2} 2x} dx$$

$$=\int \dfrac{4\sec^{2} 2x}{3\sec^{2} 2x+1} dx$$ since, $$\sec 2x\times\cos 2x =1$$

$$=\int \dfrac{4\sec^{2} 2x}{3\left( 1+\tan^{2} 2x\right) +1} dx$$ $$\because \sec^{2} \theta =1+\tan^{2} \theta$$

$$=\int \dfrac{4\sec^{2} 2x}{3+3\tan^{2} 2x+1} dx$$

$$=\int \dfrac{4\sec^{2} 2x}{4+3\tan^{2} 2x} dx$$

Now we are going to use the substitution method,

Let, $$\tan 2x=z$$

Now differentiating both side w.r.t ‘x’ we get,

$$2\sec^{2} 2xdx=dz$$

$$\sec^{2} 2xdx=\dfrac{1}{2} dz$$

Now substituting the value of $$\tan 2x$$ we get,

$$I=\int \dfrac{4\left( \dfrac{1}{2} \right) dz}{4+3z^{2}}$$

$$=\int \dfrac{2dz}{3z^{2}+4}$$

taking 3 common from denominator, we get,

$$=\dfrac{2}{3} \int \dfrac{dz}{z^{2}+\dfrac{4}{3} }$$

$$=\dfrac{2}{3} \int \dfrac{dz}{z^{2}+\left( \dfrac{2}{\sqrt{3} } \right)^{2} }$$

The above integration is in the form of $$\int \dfrac{dx}{x^{2}+a^{2}}$$,

And as we know that $$\int \dfrac{dx}{x^{2}+a^{2}} =\dfrac{1}{a} \tan^{-1} \left( \dfrac{x}{a} \right) +c$$

So by using the above formula we can write,

$$I=\dfrac{2}{3} \times \dfrac{1}{\left( \dfrac{2}{\sqrt{3} } \right) } \tan^{-1} \left( \dfrac{z}{\left( \dfrac{2}{\sqrt{3} } \right) } \right) +c$$

$$=\dfrac{2}{3} \times \dfrac{\sqrt{3} }{2} \tan^{-1} \left( \dfrac{\sqrt{3} z}{2} \right) +c$$

$$=\dfrac{1}{\sqrt{3} } \tan^{-1} \left( \dfrac{\sqrt{3} \tan 2x}{2} \right) +c (since, z=\tan 2x) $$

Which is our required solution.

Note: To solve this type of question you need to know that if the derivative of any term of denominator is present in the numerator then only you can apply the substitution method, here in order to get that derivative in numerator we have multiplied $$\sec^{2} 2x$$ in the numerator and denominator.