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Indefinite Integrals question

2019 · 12 Apr · Shift 2 · Q26
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Indefinite Integrals question

2019 · 12 Apr · Shift 2 · Q26

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
Let a∈(0,π2)a \in \left( {0,{\pi \over 2}} \right)a∈(0,2π​) be fixed. If the integral ∫tan⁡x+tan⁡αtan⁡x−tan⁡αdx\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx∫tanx−tanαtanx+tanα​dx = A(x) cos 2 α\alphaα + B(x) sin 2 α\alphaα + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively :
  1. A
    x−αx - \alphax−α and log⁡e∣cos⁡(x−α)∣{\log _e}\left| {\cos \left( {x - \alpha } \right)} \right|loge​∣cos(x−α)∣
  2. B
    x+αx + \alphax+α and log⁡e∣sin⁡(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|loge​∣sin(x−α)∣
  3. C
    x+αx + \alphax+α and log⁡e∣sin⁡(x+α)∣{\log _e}\left| {\sin \left( {x + \alpha } \right)} \right|loge​∣sin(x+α)∣
  4. D
    x−αx - \alphax−α and log⁡e∣sin⁡(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|loge​∣sin(x−α)∣
View written solutionFree

Correct answer: D

  1. Start with the given integral

We need to evaluate I=∫tan⁡x+tan⁡αtan⁡x−tan⁡α dx,I=\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha}\,dx,I=∫tanx−tanαtanx+tanα​dx, where α∈(0,π2)\alpha\in\left(0,\frac\pi2\right)α∈(0,2π​) is fixed.

We want it in the form I=A(x)cos⁡2α+B(x)sin⁡2α+C.I=A(x)\cos 2\alpha + B(x)\sin 2\alpha + C.I=A(x)cos2α+B(x)sin2α+C.

  1. Rewrite the integrand using sine and cosine

Using tan⁡x=sin⁡xcos⁡x,tan⁡α=sin⁡αcos⁡α,\tan x=\frac{\sin x}{\cos x},\qquad \tan \alpha=\frac{\sin \alpha}{\cos \alpha},tanx=cosxsinx​,tanα=cosαsinα​, we get

=\frac{\frac{\sin x\cos\alpha+\cos x\sin\alpha}{\cos x\cos\alpha}}{\frac{\sin x\cos\alpha-\cos x\sin\alpha}{\cos x\cos\alpha}} =\frac{\sin(x+\alpha)}{\sin(x-\alpha)}.$$ So $$I=\int \frac{\sin(x+\alpha)}{\sin(x-\alpha)}\,dx.$$ --- 3. **Expand $\sin(x+\alpha)$ in terms of $\sin(x-\alpha)$** Use $$\sin(x+\alpha)=\sin[(x-\alpha)+2\alpha].$$ Then $$\sin(x+\alpha)=\sin(x-\alpha)\cos 2\alpha+\cos(x-\alpha)\sin 2\alpha.$$ Therefore, $$\frac{\sin(x+\alpha)}{\sin(x-\alpha)} =\cos 2\alpha+\sin 2\alpha\,\cot(x-\alpha).$$ Hence $$I=\int \left[\cos 2\alpha+\sin 2\alpha\,\cot(x-\alpha)\right]dx.$$ --- 4. **Integrate term-by-term** Since $\alpha$ is constant, $$I=\cos 2\alpha\int dx+\sin 2\alpha\int \cot(x-\alpha)\,dx.$$ Now, $$\int dx=x,$$ and $$\int \cot(x-\alpha)\,dx=\ln|\sin(x-\alpha)|.$$ So, $$I=x\cos 2\alpha+\ln|\sin(x-\alpha)|\sin 2\alpha+C.$$ --- 5. **Match with the required form** Comparing with $$I=A(x)\cos 2\alpha+B(x)\sin 2\alpha+C,$$ we get $$A(x)=x, \qquad B(x)=\ln|\sin(x-\alpha)|.$$ But the options contain $x-\alpha$ instead of $x$. This is acceptable because $\alpha$ is a constant, so $$(x-\alpha)\cos 2\alpha = x\cos 2\alpha - \alpha\cos 2\alpha,$$ and the extra term $-\alpha\cos 2\alpha$ is just a constant and can be absorbed into $C$. Thus we may write equivalently, $$I=(x-\alpha)\cos 2\alpha+\ln|\sin(x-\alpha)|\sin 2\alpha+C.$$ So, $$A(x)=x-\alpha, \qquad B(x)=\ln|\sin(x-\alpha)|.$$ --- 6. **Check options** - **A:** $A=x-\alpha$, $B=\ln|\cos(x-\alpha)|$ ❌ - **B:** $A=x+\alpha$, $B=\ln|\sin(x-\alpha)|$ ❌ - **C:** $A=x+\alpha$, $B=\ln|\sin(x+\alpha)|$ ❌ - **D:** $A=x-\alpha$, $B=\ln|\sin(x-\alpha)|$ ✅ Therefore, the correct option is **D**.
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