Class 12 Integration: 3 Sure Questions

This lesson solves three frequently asked Class 12 integration questions, building each answer step by step with identities, substitution, and integration by parts.

Question 1: integrating $\dfrac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x}$

We want to evaluate

$I = \int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \, \cos^2 x} \, dx.$

Write the numerator as a sum of cubes, $(\sin^2 x)^3 + (\cos^2 x)^3$, and use $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$ with $a = \sin^2 x$ and $b = \cos^2 x$:

$\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x (\sin^2 x + \cos^2 x).$

Since $\sin^2 x + \cos^2 x = 1$, this collapses to

$\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x.$

Dividing through by $\sin^2 x \cos^2 x$ gives

$I = \int \left( \frac{1}{\sin^2 x \cos^2 x} - 3 \right) dx.$

Replacing the $1$ in the first term with $\sin^2 x + \cos^2 x$ splits it into two standard integrals:

$\frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \sec^2 x + \csc^2 x.$

Therefore

$I = \int \sec^2 x \, dx + \int \csc^2 x \, dx - \int 3 \, dx = \tan x - \cot x - 3x + C.$

Question 2: integrating $\dfrac{\sin^{-1} x}{(1 - x^2)^{3/2}}$

We want

$I = \int \frac{\sin^{-1} x}{(1 - x^2)^{3/2}} \, dx = \int \frac{\sin^{-1} x}{(1 - x^2)\sqrt{1 - x^2}} \, dx.$

Substitution

Put $\sin^{-1} x = t$, so $x = \sin t$ and $\dfrac{1}{\sqrt{1 - x^2}} \, dx = dt$. Then $1 - x^2 = 1 - \sin^2 t = \cos^2 t$, and the integral becomes

$I = \int \frac{t}{1 - \sin^2 t} \, dt = \int \frac{t}{\cos^2 t} \, dt = \int t \sec^2 t \, dt.$

Integration by parts

Take the first function $t$ and the second function $\sec^2 t$:

$I = t \int \sec^2 t \, dt - \int \left( \frac{d}{dt}(t) \int \sec^2 t \, dt \right) dt = t \tan t - \int \tan t \, dt.$

Since $\int \tan t \, dt = -\log|\cos t|$,

$I = t \tan t + \log|\cos t| + C.$

Back to $x$

With $t = \sin^{-1} x$, $\tan t = \dfrac{\sin t}{\cos t} = \dfrac{x}{\sqrt{1 - x^2}}$ and $\cos t = \sqrt{1 - x^2}$, so

$I = \frac{x \, \sin^{-1} x}{\sqrt{1 - x^2}} + \log\left|\sqrt{1 - x^2}\right| + C.$

Question 3: integrating $\sqrt{\dfrac{a + x}{a - x}}$

We want

$I = \int \sqrt{\frac{a + x}{a - x}} \, dx.$

Multiply numerator and denominator inside the root by the conjugate $a + x$:

$I = \int \sqrt{\frac{(a + x)^2}{(a - x)(a + x)}} \, dx = \int \frac{a + x}{\sqrt{a^2 - x^2}} \, dx.$

Split into two integrals, $I = I_1 + I_2$, where

$I_1 = \int \frac{a}{\sqrt{a^2 - x^2}} \, dx, \qquad I_2 = \int \frac{x}{\sqrt{a^2 - x^2}} \, dx.$

First part. Using $\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\dfrac{x}{a}$,

$I_1 = a \, \sin^{-1}\frac{x}{a} + C_1.$

Second part. Put $a^2 - x^2 = t$, so $-2x \, dx = dt$, that is $x \, dx = -\tfrac{1}{2} \, dt$:

$I_2 = \int \frac{-\tfrac{1}{2} \, dt}{\sqrt{t}} = -\tfrac{1}{2} \cdot 2\sqrt{t} = -\sqrt{a^2 - x^2} + C_2.$

Adding the two parts and combining the constants,

$I = a \, \sin^{-1}\frac{x}{a} - \sqrt{a^2 - x^2} + C.$

Key takeaways

• $\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x$, so $\displaystyle\int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x}\,dx = \tan x - \cot x - 3x + C$.
• For $\dfrac{\sin^{-1} x}{(1 - x^2)^{3/2}}$, substitute $\sin^{-1} x = t$ to reach $\int t \sec^2 t \, dt$, then integrate by parts.
• A surd of the form $\sqrt{\tfrac{a+x}{a-x}}$ rationalises to $\dfrac{a + x}{\sqrt{a^2 - x^2}}$, which splits into a standard inverse sine integral and a simple substitution.