CUET UG 2025 — Mathematics Calculus

The integral to evaluate: $\displaystyle \int \frac{\sin 2x \, dx}{\sqrt{9 - \cos^4 x}}$∫9−cos4x​sin2xdx​

Using the double angle formula:

$$\sin 2x = 2\sin x \cos x$$sin2x=2sinxcosx

The integral becomes:

$$\int \frac{2\sin x \cos x \, dx}{\sqrt{9 - \cos^4 x}}$$∫9−cos4x​2sinxcosxdx​

---

Let $u = \cos^2 x$u=cos2x

$$\frac{du}{dx} = 2\cos x \cdot (-\sin x)$$dxdu​=2cosx⋅(−sinx)

$$\frac{du}{dx} = -2\sin x \cos x$$dxdu​=−2sinxcosx

$$du = -2\sin x \cos x \, dx$$du=−2sinxcosxdx

$$2\sin x \cos x \, dx = -du$$2sinxcosxdx=−du

Also, $\cos^4 x = (\cos^2 x)^2 = u^2$cos4x=(cos2x)2=u2

---

Substituting into the integral:

$$\int \frac{-du}{\sqrt{9 - u^2}}$$∫9−u2​−du​

$$= -\int \frac{du}{\sqrt{9 - u^2}}$$=−∫9−u2​du​

---

Using the standard integral formula $\displaystyle \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$∫a2−x2​dx​=sin−1(ax​)+C

Here $a^2 = 9$a2=9, so $a = 3$a=3

$$-\int \frac{du}{\sqrt{9 - u^2}} = -\sin^{-1}\left(\frac{u}{3}\right) + C$$−∫9−u2​du​=−sin−1(3u​)+C

---

Substituting back $u = \cos^2 x$u=cos2x:

$$-\sin^{-1}\left(\frac{\cos^2 x}{3}\right) + C$$−sin−1(3cos2x​)+C