Question

Find the solution of the integral $\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}$ ?

Answer 1

We have to integrate $\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}$with respect to ‘x’. We solve this using a substitution method. We simplify the integrand using the formula $\sin 2x = 2\sin x.\cos x$ and by dividing the numerator and the denominator by ${\cos ^4}x$. We also know the tangent function is the ratio of sine function to cosine function. After simplification we use the substitution method and we put the final answer in terms of ‘x’ only.

Complete step by step answer:
Given $\int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx$
Now we know that $\sin 2x = 2\sin x.\cos x$
$\Rightarrow \int {\dfrac{{2\sin x.\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx$
Divide the numerator and the denominator by ${\cos ^4}x$
$\int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx= \int {\dfrac{{\left( {\dfrac{{2\sin x.\cos x}}{{{{\cos }^4}x}}} \right)}}{{\left( {\dfrac{{{{\sin }^4}x + {{\cos }^4}x}}{{{{\cos }^4}x}}} \right)}}} dx$
Splitting the terms in denominator,
$\int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx= \int {\dfrac{{\left( {\dfrac{{2\sin x.\cos x}}{{{{\cos }^4}x}}} \right)}}{{\left( {\dfrac{{{{\sin }^4}x}}{{{{\cos }^4}x}} + \dfrac{{{{\cos }^4}x}}{{{{\cos }^4}x}}} \right)}}} dx$

Cancelling we have
$\int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx = \int {\dfrac{{\left( {\dfrac{{2\sin x}}{{{{\cos }^3}x}}} \right)}}{{\left( {\dfrac{{{{\sin }^4}x}}{{{{\cos }^4}x}} + 1} \right)}}} dx$
$\Rightarrow \int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx = 2\int {\dfrac{{\left( {\dfrac{{\sin x}}{{\cos x}} \times \dfrac{1}{{{{\cos }^2}x}}} \right)}}{{\left( {\dfrac{{{{\sin }^4}x}}{{{{\cos }^4}x}} + 1} \right)}}} dx$
We know that secant and cosecant are reciprocal function and using the definition of tangent function we have,
$\int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx = \int {\dfrac{{2\left( {\tan x.{{\sec }^2}x} \right)}}{{\left( {{{\tan }^4}x + 1} \right)}}} dx$
Now to simplify thus easily we use a substitute method.
Let’s put $t = {\tan ^2}x$then differentiating we have
$dt = 2\tan x{\sec ^2}xdx$

Then above integral becomes
$\int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx = \int {\dfrac{{dt}}{{\left( {{t^2} + 1} \right)}}}$
$\Rightarrow \int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx= \int {\dfrac{{dt}}{{\left( {1 + {t^2}} \right)}}}$
But we know the integration of $\int {\dfrac{{dx}}{{1 + {x^2}}}} = {\tan ^{ - 1}}x + c$
$\int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx= {\tan ^{ - 1}}(t) + c$
Where ‘c’ is the integration constant.
But we have substituted $t = {\tan ^2}x$, then we have
$\therefore \int {\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}} dx = {\tan ^{ - 1}}({\tan ^2}x) + c$

Thus, the integration of $\dfrac{{\sin 2x}}{{{{\sin }^4}x + {{\cos }^4}x}}$ is ${\tan ^{ - 1}}({\tan ^2}x) + c$, where ‘c’ is integration constant.

Note: We have indefinite integral that is why we added integration constant. If we have a definite integral we do not add integration constant. We know the integration formulas which yield inverse trigonometric functions:
$\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {{\sin }^{ - 1}}\left( {\dfrac{x}{a}} \right)} + c$
$\Rightarrow \int {\dfrac{{dx}}{{{a^2} + {x^2}}} = {{\tan }^{ - 1}}\left( {\dfrac{x}{a}} \right)} + c$
$\Rightarrow \int {\dfrac{{dx}}{{x\sqrt {{x^2} - {a^2}} }} = \dfrac{1}{a}{{\sec }^{ - 1}}\left( {\dfrac{{|x|}}{a}} \right)} + c$
We use them according to the given problem.