If \int \tan x \over 1 + \tan x + \tan ^2xdx = x - K \over \sqrt A \tan ^ - 1 ( K\,\tan x + 1 \over \sqrt A ) + C,(C\,\, is a constant of integration) then the ordered pair (K, A) is equal to :

Indefinite Integrals PYQ — JEE Main 2018 · 16 Apr · Shift 1 · Q27 with Solution

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Correct answer: C

Evaluate the integral

We need to compute $I=\int \frac{\tan x}{1+\tan x+\tan^2 x}\,dx.$

Let $t=\tan x \implies dx=\frac{dt}{1+t^2}.$ Then $I=\int \frac{t}{1+t+t^2}\cdot \frac{dt}{1+t^2}$ which is not the most convenient route.

A better way is to rewrite the integrand directly in terms of $\sin x,\cos x$ :

=\frac{\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}+\frac{\sin^2 x}{\cos^2 x}}.$$ Multiply numerator and denominator by $\cos^2 x$: $$=\frac{\sin x\cos x}{\cos^2 x+\sin x\cos x+\sin^2 x}.$$ Using $\sin^2 x+\cos^2 x=1$, $$\cos^2 x+\sin x\cos x+\sin^2 x=1+\sin x\cos x.$$ So $$I=\int \frac{\sin x\cos x}{1+\sin x\cos x}\,dx.$$ This still does not directly match the given form. So instead, let us simplify the original integrand algebraically: Write $$\frac{t}{t^2+t+1}=\frac12\cdot\frac{2t+1}{t^2+t+1}-\frac12\cdot\frac1{t^2+t+1}.$$ Hence with $t=\tan x$, $$I=\int \frac{\tan x}{1+\tan x+\tan^2 x}\,dx.$$ Now compare with the form given in the question: $$I=x-\frac{K}{\sqrt A}\tan^{-1}\left(\frac{K\tan x+1}{\sqrt A}\right)+C.$$ So differentiate the RHS and match the integrand. 2. **Differentiate the given expression** Let $$F(x)=x-\frac{K}{\sqrt A}\tan^{-1}\left(\frac{K\tan x+1}{\sqrt A}\right).$$ Then $$F'(x)=1-\frac{K}{\sqrt A}\cdot \frac{1}{1+\left(\frac{K\tan x+1}{\sqrt A}\right)^2}\cdot \frac{K\sec^2 x}{\sqrt A}.$$ Thus $$F'(x)=1-\frac{K^2\sec^2 x}{A+(K\tan x+1)^2}.$$ Now expand denominator: $$(K\tan x+1)^2=K^2\tan^2 x+2K\tan x+1,$$ so $$F'(x)=1-\frac{K^2(1+\tan^2 x)}{A+K^2\tan^2 x+2K\tan x+1}.$$ Combine into one fraction: $$F'(x)=\frac{A+K^2\tan^2 x+2K\tan x+1-K^2-K^2\tan^2 x}{A+K^2\tan^2 x+2K\tan x+1}.$$ Therefore $$F'(x)=\frac{(A+1-K^2)+2K\tan x}{A+1+2K\tan x+K^2\tan^2 x}.$$ We want this to equal $$\frac{\tan x}{1+\tan x+\tan^2 x}.$$ 3. **Match coefficients** For exact equality, numerator and denominator must be proportional in the same way. Since the target numerator is just $\tan x$, we need $$A+1-K^2=0$$ and $$2K\tan x \propto \tan x.$$ Also denominator should match $$1+\tan x+\tan^2 x.$$ So comparing $$A+1+2K\tan x+K^2\tan^2 x$$ with $$1+\tan x+\tan^2 x,$$ we need $$K^2=1\quad ?$$ This seems inconsistent with the options, so let us test the options directly. 4. **Test option C: $(K,A)=(2,3)$** Then $$F(x)=x-\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2\tan x+1}{\sqrt3}\right).$$ Differentiate: $$F'(x)=1-\frac{2}{\sqrt3}\cdot \frac{1}{1+\left(\frac{2\tan x+1}{\sqrt3}\right)^2}\cdot \frac{2\sec^2 x}{\sqrt3}.$$ So $$F'(x)=1-\frac{4\sec^2 x}{3+(2\tan x+1)^2}.$$ Now $$3+(2\tan x+1)^2=3+4\tan^2 x+4\tan x+1=4(1+\tan x+\tan^2 x).$$ Hence $$F'(x)=1-\frac{4\sec^2 x}{4(1+\tan x+\tan^2 x)} =1-\frac{1+\tan^2 x}{1+\tan x+\tan^2 x}.$$ Therefore $$F'(x)=\frac{1+\tan x+\tan^2 x-(1+\tan^2 x)}{1+\tan x+\tan^2 x} =\frac{\tan x}{1+\tan x+\tan^2 x}.$$ This matches the integrand exactly. So option C is correct. 5. **Check other options briefly** - For $(2,1)$, denominator becomes $1+(2\tan x+1)^2=4\tan^2 x+4\tan x+2$, which does not simplify to a multiple of $1+\tan x+\tan^2 x$ appropriately. - For $(-2,3)$, differentiation gives numerator with $-\tan x$, wrong sign. - For $(-2,1)$, also incorrect. Therefore the ordered pair is $$\boxed{(K,A)=(2,3)}.$$

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View written solutionFree

Correct answer: C

Evaluate the integral

We need to compute $I=\int \frac{\tan x}{1+\tan x+\tan^2 x}\,dx.$

Let $t=\tan x \implies dx=\frac{dt}{1+t^2}.$ Then $I=\int \frac{t}{1+t+t^2}\cdot \frac{dt}{1+t^2}$ which is not the most convenient route.

A better way is to rewrite the integrand directly in terms of $\sin x,\cos x$ :

=\frac{\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}+\frac{\sin^2 x}{\cos^2 x}}.$$ Multiply numerator and denominator by $\cos^2 x$: $$=\frac{\sin x\cos x}{\cos^2 x+\sin x\cos x+\sin^2 x}.$$ Using $\sin^2 x+\cos^2 x=1$, $$\cos^2 x+\sin x\cos x+\sin^2 x=1+\sin x\cos x.$$ So $$I=\int \frac{\sin x\cos x}{1+\sin x\cos x}\,dx.$$ This still does not directly match the given form. So instead, let us simplify the original integrand algebraically: Write $$\frac{t}{t^2+t+1}=\frac12\cdot\frac{2t+1}{t^2+t+1}-\frac12\cdot\frac1{t^2+t+1}.$$ Hence with $t=\tan x$, $$I=\int \frac{\tan x}{1+\tan x+\tan^2 x}\,dx.$$ Now compare with the form given in the question: $$I=x-\frac{K}{\sqrt A}\tan^{-1}\left(\frac{K\tan x+1}{\sqrt A}\right)+C.$$ So differentiate the RHS and match the integrand. 2. **Differentiate the given expression** Let $$F(x)=x-\frac{K}{\sqrt A}\tan^{-1}\left(\frac{K\tan x+1}{\sqrt A}\right).$$ Then $$F'(x)=1-\frac{K}{\sqrt A}\cdot \frac{1}{1+\left(\frac{K\tan x+1}{\sqrt A}\right)^2}\cdot \frac{K\sec^2 x}{\sqrt A}.$$ Thus $$F'(x)=1-\frac{K^2\sec^2 x}{A+(K\tan x+1)^2}.$$ Now expand denominator: $$(K\tan x+1)^2=K^2\tan^2 x+2K\tan x+1,$$ so $$F'(x)=1-\frac{K^2(1+\tan^2 x)}{A+K^2\tan^2 x+2K\tan x+1}.$$ Combine into one fraction: $$F'(x)=\frac{A+K^2\tan^2 x+2K\tan x+1-K^2-K^2\tan^2 x}{A+K^2\tan^2 x+2K\tan x+1}.$$ Therefore $$F'(x)=\frac{(A+1-K^2)+2K\tan x}{A+1+2K\tan x+K^2\tan^2 x}.$$ We want this to equal $$\frac{\tan x}{1+\tan x+\tan^2 x}.$$ 3. **Match coefficients** For exact equality, numerator and denominator must be proportional in the same way. Since the target numerator is just $\tan x$, we need $$A+1-K^2=0$$ and $$2K\tan x \propto \tan x.$$ Also denominator should match $$1+\tan x+\tan^2 x.$$ So comparing $$A+1+2K\tan x+K^2\tan^2 x$$ with $$1+\tan x+\tan^2 x,$$ we need $$K^2=1\quad ?$$ This seems inconsistent with the options, so let us test the options directly. 4. **Test option C: $(K,A)=(2,3)$** Then $$F(x)=x-\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2\tan x+1}{\sqrt3}\right).$$ Differentiate: $$F'(x)=1-\frac{2}{\sqrt3}\cdot \frac{1}{1+\left(\frac{2\tan x+1}{\sqrt3}\right)^2}\cdot \frac{2\sec^2 x}{\sqrt3}.$$ So $$F'(x)=1-\frac{4\sec^2 x}{3+(2\tan x+1)^2}.$$ Now $$3+(2\tan x+1)^2=3+4\tan^2 x+4\tan x+1=4(1+\tan x+\tan^2 x).$$ Hence $$F'(x)=1-\frac{4\sec^2 x}{4(1+\tan x+\tan^2 x)} =1-\frac{1+\tan^2 x}{1+\tan x+\tan^2 x}.$$ Therefore $$F'(x)=\frac{1+\tan x+\tan^2 x-(1+\tan^2 x)}{1+\tan x+\tan^2 x} =\frac{\tan x}{1+\tan x+\tan^2 x}.$$ This matches the integrand exactly. So option C is correct. 5. **Check other options briefly** - For $(2,1)$, denominator becomes $1+(2\tan x+1)^2=4\tan^2 x+4\tan x+2$, which does not simplify to a multiple of $1+\tan x+\tan^2 x$ appropriately. - For $(-2,3)$, differentiation gives numerator with $-\tan x$, wrong sign. - For $(-2,1)$, also incorrect. Therefore the ordered pair is $$\boxed{(K,A)=(2,3)}.$$

Indefinite Integrals question

Indefinite Integrals question

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