$Evaluate : \int \frac{\sin x + \cos x}{16 + 9 \sin 2 x} dx$

Solution

$Let I = \int \frac{\sin x + \cos x}{16 + 9 2 x} dx$

Here, we express the denominator in terms of sin x - cos x which is the integration of numerator.

Clearly,

$(\sin x - \cos x)^{2} = \sin^{2} x + \cos^{2} x - 2 \sin xcos x = 1 - \sin 2 x \Rightarrow \sin 2 x = 1 - (\sin x - \cos x)^{2} ∴ I = \int_{0}^{π / 4} \frac{\sin x + \cos x}{16 + 9 \{1 - (\sin x - \cos x)^{2}\}} dx \Rightarrow I = \int_{0}^{π / 4} \frac{\sin x + \cos x}{25 - 9 (\sin x - \cos x)^{2}} dx Let \sin x - cosx = t, Then, d (\sin x - \cos x) = dt \Rightarrow (\cos x + \sin x) dx = dt also, x = 0 \Rightarrow t = \sin 0 - \cos 0 = - 1 and x = \frac{π}{4} t = \sin \frac{π}{4} - \cos \frac{π}{4} = 0 ∴ I = \int_{- 1}^{0} \frac{dt}{25 - 9 t^{2}} = \frac{1}{9} \int_{- 1}^{0} \frac{dt}{\frac{25}{9} - t^{2}} = \frac{1}{9} \int_{- 1}^{0} \frac{dt}{{(\frac{5}{3})}^{2} - t^{2}} \Rightarrow I = \frac{I}{9} x \frac{1}{2 (5 / 3)} {[\log |\frac{\frac{5}{3} + t}{5 / 3 - t}|]}_{- 1}^{0} \Rightarrow I = \frac{1}{30} [\log - 1 \log (\frac{2 / 3}{8 / 3})] = \frac{1}{30} [\log 1 - \log (\frac{1}{4})] = \frac{1}{30} [\log 1 + \log 4] = \frac{1}{30} \log 4 = \frac{1}{15} \log 2$

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$Evaluate : \int \frac{\sin x + \cos x}{16 + 9 \sin 2 x} dx$

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Integrals

Evaluate: ∫sin x + cos x16 + 9 sin 2x dx

Some More Questions From Integrals Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

$Evaluate : \int \frac{\sin x + \cos x}{16 + 9 \sin 2 x} dx$