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Class 12Calculus7:43Published 23 Sept 2025

Integration: Root of (1 - root x)/(1 + root x) (Misc. Exercise Q19)

A full walkthrough of Question 19 from the Chapter 7 (Integration) miscellaneous exercise: integrating the square root of (1 minus root x) over (1 plus root x) using a trigonometric substitution.

This lesson solves an important miscellaneous exercise problem from the integration chapter for Class 12. The teacher substitutes root x equal to cosine t, rewrites the integrand with half-angle identities, and reduces it to a simple integral of cosine and cosine squared. After integrating, the answer is converted back into x to give a clean closed form, with every identity used listed at the end.

What you'll learn

Choosing a trigonometric substitution to simplify a messy square-root integrand
Using half-angle identities to turn a ratio of cosines into a tangent
Integrating cosine squared by rewriting it as a double-angle expression
Converting the result back from the angle to the original variable

Lesson chapters

0:00The problem and the substitution

1:18Half-angle identities simplify the root

3:05Reducing to cosine and cosine squared

3:43Integrating cosine squared with the double angle

5:03Substituting back to x for the final answer

6:35Summary of identities used

Lesson notes

This lesson works through Question 19 of the Chapter 7 (Integration) miscellaneous exercise: evaluating $\displaystyle \int \sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx$ . The key idea is a trigonometric substitution that clears the nested square roots.

The substitution

Let $\sqrt{x} = \cos t$ , so that $t = \cos^{-1}\sqrt{x}$ .

Squaring gives $x = \cos^2 t$ . Differentiating,

$dx = -2\cos t\,\sin t\,dt.$

Substituting $\sqrt{x} = \cos t$ into the integrand,

$I = \int \sqrt{\frac{1-\cos t}{1+\cos t}}\,(-2\cos t\,\sin t)\,dt.$

Simplifying with half-angle identities

Use $1-\cos t = 2\sin^2\tfrac{t}{2}$ and $1+\cos t = 2\cos^2\tfrac{t}{2}$ , so the root becomes

$\sqrt{\frac{1-\cos t}{1+\cos t}} = \sqrt{\frac{2\sin^2\tfrac{t}{2}}{2\cos^2\tfrac{t}{2}}} = \frac{\sin\tfrac{t}{2}}{\cos\tfrac{t}{2}}.$

Also write $\sin t = 2\sin\tfrac{t}{2}\cos\tfrac{t}{2}$ . Then

$I = -2\int \frac{\sin\tfrac{t}{2}}{\cos\tfrac{t}{2}}\,\big(2\sin\tfrac{t}{2}\cos\tfrac{t}{2}\big)\cos t\,dt = -4\int \sin^2\tfrac{t}{2}\,\cos t\,dt.$

Reducing to cosine and cosine squared

Replace $\sin^2\tfrac{t}{2} = \dfrac{1-\cos t}{2}$ :

$I = -4\int \frac{1-\cos t}{2}\,\cos t\,dt = -2\int (1-\cos t)\cos t\,dt = -2\int \big(\cos t - \cos^2 t\big)\,dt.$

Integrating cosine squared

The term $\cos^2 t$ is rewritten with the double-angle identity

$\cos^2 t = \frac{1+\cos 2t}{2}.$

Integrating term by term,

$\int \cos t\,dt = \sin t, \qquad \int \cos^2 t\,dt = \frac{t}{2} + \frac{\sin 2t}{4}.$

So

$I = -2\left[\sin t - \frac{t}{2} - \frac{\sin 2t}{4}\right] = -2\sin t + t + \frac{\sin 2t}{2}.$

Substituting back to x

Recall $t = \cos^{-1}\sqrt{x}$ . Then

$\sin t = \sin\!\big(\cos^{-1}\sqrt{x}\big) = \sqrt{1-x}, \qquad \cos t = \sqrt{x},$

and $\sin 2t = 2\sin t\cos t = 2\sqrt{1-x}\,\sqrt{x}$ . Putting these in,

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

Key takeaways

The substitution $\sqrt{x} = \cos t$ clears the nested square roots and turns the integrand into trig functions.
Half-angle identities collapse $\sqrt{\tfrac{1-\cos t}{1+\cos t}}$ to $\tan\tfrac{t}{2}$ , and the double-angle identity handles $\cos^2 t$ .
The final answer is $\displaystyle -2\sqrt{1-x} + \cos^{-1}\sqrt{x} + \sqrt{x}\,\sqrt{1-x} + C$ .