Inspired by Harsh Shrivastava

Calculus Level 4

If $\tan(\beta)=\dfrac{n \sin(\omega) \cos (\omega)}{1-n \sin^{2} (\omega)}$

Find the value of $\large\int\limits_0^{\pi/4} \frac{\tan(\omega-\beta)}{1-n}\,\mathrm d\omega$

Answer up to to three decimal places.

2 solutions

Tanishq Varshney
Dec 16, 2015

On multiplying and dividing by $\cos^2 (\omega)$

$\large{\tan (\beta)=\frac{n \tan (\omega)}{\sec^2 (\omega)-n \tan^2 (\omega)}}$

$\large{\tan (\beta)=\frac{n \tan (\omega)}{1+(1-n) \tan^2 (\omega)}}$

if we place n=2 and recall identity of $\tan 2x$

Thus we have $\tan (\beta)=\tan (2 \omega)$ or $\beta =2 \omega$

Now we have to integrate $\large{\frac{\tan(\omega - 2 \omega )}{1-2}}$ ,in short $\tan (\omega)$

$\large{\ln(\sec \omega) |_{0}^{\frac{\pi}{4}}}$

The final answer is $\large{\frac{1}{2} \ln (2)}$

Pulkit Gupta
Dec 17, 2015

Clearly, the answer does not depend on the value of n taken. So we put n = 0 to obtain $\tan \beta$ = 0.

The integral then reduces to $\tan \omega$ , which is equal to $\large{\ln(\sec \omega) |_{0}^{\frac{\pi}{4}}}$ in the given question.

Why the answer is not depending on n? Could you explain

Eziz Hudaykulyyev - 5 years, 5 months ago

Had the answer would have depended on n, the answer should not have come as a constant value :)

Pulkit Gupta - 5 years, 5 months ago