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Definite Integration question

2015 · Shift 2 · Q39
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Definite Integration question

2015 · Shift 2 · Q39

JEE AdvancedMathematicsDefinite IntegrationNumerical+4 / −1
If α=∫01(e9x+3tan⁡−1x)(12+9x21+x2)dx\alpha = \int\limits_0^1 {\left( {{e^{9x + 3{{\tan }^{ - 1}}x}}} \right)\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)} dxα=0∫1​(e9x+3tan−1x)(1+x212+9x2​)dx where tan⁡−1x{\tan ^{ - 1}}xtan−1x takes only principal values, then the value of (log⁡e∣1+α∣−3π4)\left( {{{\log }_e}\left| {1 + \alpha } \right| - {{3\pi } \over 4}} \right)(loge​∣1+α∣−43π​) is
Numerical answer
View written solutionFree

Correct answer: 9

  1. We need to evaluate
α=∫01e9x+3tan⁡−1x(12+9x21+x2)dx.\alpha=\int_0^1 e^{9x+3\tan^{-1}x}\left(\frac{12+9x^2}{1+x^2}\right)dx.α=∫01​e9x+3tan−1x(1+x212+9x2​)dx.
  1. Look for a substitution using the exponent:
f(x)=9x+3tan⁡−1x.f(x)=9x+3\tan^{-1}x.f(x)=9x+3tan−1x.

Then

f′(x)=9+31+x2=9(1+x2)+31+x2=12+9x21+x2.f'(x)=9+\frac{3}{1+x^2}= \frac{9(1+x^2)+3}{1+x^2}= \frac{12+9x^2}{1+x^2}.f′(x)=9+1+x23​=1+x29(1+x2)+3​=1+x212+9x2​.

This matches the remaining factor in the integrand.

So the integral becomes

α=∫01ef(x)f′(x) dx.\alpha=\int_0^1 e^{f(x)}f'(x)\,dx.α=∫01​ef(x)f′(x)dx.

Hence,

α=ef(x)∣01.\alpha=e^{f(x)}\Big|_0^1.α=ef(x)​01​.
  1. Compute the limits:

At x=1x=1x=1,

f(1)=9(1)+3tan⁡−1(1)=9+3⋅π4=9+3π4.f(1)=9(1)+3\tan^{-1}(1)=9+3\cdot \frac{\pi}{4}=9+\frac{3\pi}{4}.f(1)=9(1)+3tan−1(1)=9+3⋅4π​=9+43π​.

At x=0x=0x=0,

f(0)=0+3tan⁡−1(0)=0.f(0)=0+3\tan^{-1}(0)=0.f(0)=0+3tan−1(0)=0.

Therefore,

α=e9+3π4−e0=e9+3π4−1.\alpha=e^{9+\frac{3\pi}{4}}-e^0=e^{9+\frac{3\pi}{4}}-1.α=e9+43π​−e0=e9+43π​−1.
  1. Now compute
log⁡e∣1+α∣−3π4.\log_e|1+\alpha|-\frac{3\pi}{4}.loge​∣1+α∣−43π​.

Since

1+α=1+(e9+3π4−1)=e9+3π4>0,1+\alpha=1+\left(e^{9+\frac{3\pi}{4}}-1\right)=e^{9+\frac{3\pi}{4}}>0,1+α=1+(e9+43π​−1)=e9+43π​>0,

we have

log⁡e∣1+α∣=log⁡e(e9+3π4)=9+3π4.\log_e|1+\alpha|=\log_e\left(e^{9+\frac{3\pi}{4}}\right)=9+\frac{3\pi}{4}.loge​∣1+α∣=loge​(e9+43π​)=9+43π​.

Thus,

log⁡e∣1+α∣−3π4=9+3π4−3π4=9.\log_e|1+\alpha|-\frac{3\pi}{4}= 9+\frac{3\pi}{4}-\frac{3\pi}{4}=9.loge​∣1+α∣−43π​=9+43π​−43π​=9.
  1. Final answer:
9\boxed{9}9​
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