Jamia Millia Islamia Indefinite Integration Previous Year Questions (PYQs) – Page 1 of 2

Let $x = 2\tan\theta$, $dx = 2\sec^2\theta,d\theta$ $\Rightarrow I = \int \dfrac{2\tan\theta \cdot 2\sec^2\theta}{2\sec\theta}d\theta = 2\int \tan\theta\sec\theta,d\theta = 2\sec\theta + C = \dfrac{1}{2}\sqrt{x^2+4}+C$ Hence equivalent to $\dfrac{1}{4}\sec^{-1}\left(\dfrac{x}{2}\right)$

$\cos A - \cos B = -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$ $\Rightarrow \cos 2x - \cos 2\theta = -2\sin(x+\theta)\sin(x-\theta)$ and $\cos x - \cos\theta = -2\sin\dfrac{x+\theta}{2}\sin\dfrac{x-\theta}{2}$ So, $\displaystyle \dfrac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta} = \dfrac{\sin(x+\theta)\sin(x-\theta)}{\sin\dfrac{x+\theta}{2}\sin\dfrac{x-\theta}{2}} = 4\cos\dfrac{x+\theta}{2}\cos\dfrac{x-\theta}{2} = 2(\cos x + \cos\theta)$ Hence, $\displaystyle \int \dfrac{\cos 2x - \cos 2\theta}{\cos x - \cos \theta},dx = \int 2(\cos x + \cos\theta),dx = 2\sin x + 2x\cos\theta + C$ $\boxed{\text{Answer: (A) }2(\sin x + x\cos\theta) + C}$

Solution: Let $t=x^3 \Rightarrow dt=3x^2dx$, so $y=\dfrac{1}{3}e^{x^3}+C$. Using $(0,1)$: $1=\dfrac{1}{3}+C \Rightarrow C=\dfrac{2}{3}$. Hence $3y-2=e^{x^3} \Rightarrow x^3=\log_e(3y-2)$, so $x=\sqrt[3]{\log_e(3y-2)}$.

Put $t=\log x$, then $dt=\frac{dx}{x}$; next $u=\log t$, $du=\frac{dt}{t}$. Integral becomes $\int \frac{du}{u}=\log u=\log(\log(\log x))+C$.

$\dfrac{d}{dx}\big(x^{x}\big)=x^{x}(1+\log x)$ (log = natural log). Hence integral $=x^{x}+C$.

Let $t=\sqrt{x}$, $dx=2t\,dt$. Then $\int 2t^{2}e^{t}dt=2e^{t}(t^{2}-2t+2)+C$. Substitute $t=\sqrt{x}$ ⇒ $(2x-4\sqrt{x}+4)e^{\sqrt{x}}+C$.

Let $u=x^{3}\Rightarrow du=3x^{2}dx\Rightarrow x^{2}dx=\tfrac{1}{3}du$. $\int x^{2}\sin(x^{3})dx=\tfrac13\int\sin u,du=-\tfrac13\cos x+C.$

Let $I = \int e^{x^2}\left(\frac{1}{x} - \frac{1}{2x^2}\right)dx$. Differentiate $e^{x^2}/x$: $\dfrac{d}{dx}\left(\dfrac{e^{x^2}}{x}\right) = e^{x^2}\left(2 - \dfrac{1}{x^2}\right)$. Thus, $I$ can be expressed as a part of $\dfrac{d}{dx}\left(\dfrac{e^{x^2}}{2x}\right)$, and on integration we get: $I = \dfrac{e^{x^2}(e^2 - 2)}{2} + C$.

Let $I = \int \frac{d\theta}{1 - \tan\theta}$. Multiply numerator and denominator by $\cos\theta$: $I = \int \frac{\cos\theta\,d\theta}{\cos\theta - \sin\theta}$. Let $u = \cos\theta - \sin\theta \Rightarrow du = -(\sin\theta + \cos\theta)d\theta$. Rewrite and integrate ⇒ $I = \dfrac{1}{2}\log|\cos\theta + \sin\theta| + C$.