0-pi-2-1-1-cos-x-2sin-x-dx-

Question Number 166612 by mnjuly1970 last updated on 23/Feb/22

\int_{0}^{\frac{π}{2}} \frac{1}{1 + c o s (x) + 2 s i n (x)} d x = ?

Commented by cortano1 last updated on 23/Feb/22

1 + \cos x + 2 \sin x = \cos^{2} \frac{1}{2} x + \sin^{2} \frac{1}{2} x + \cos^{2} \frac{1}{2} x - \sin^{2} \frac{1}{x} + 2 \sin x

= 2 \cos^{2} \frac{1}{2} x + 4 \sin \frac{1}{2} xcos \frac{1}{2} x

= 2 \cos \frac{1}{2} x (\cos \frac{1}{2} x + 2 \sin \frac{1}{2} x)

I = \frac{1}{2} \int \frac{\sec \frac{1}{2} x}{\cos \frac{1}{2} x + 2 \sin \frac{1}{2} x} dx

I = \frac{1}{2} \int \frac{\sec^{2} \frac{1}{2} x}{1 + 2 \tan \frac{1}{2} x} dx

[u = \frac{1}{2} x]

I = \int \frac{\sec^{2} u}{1 + 2 \tan u} du = \frac{1}{2} \int \frac{d (1 + 2 \tan u)}{1 + 2 \tan u}

I = \frac{1}{2} \ln ∣ 1 + 2 \tan u ∣ + c

I = \frac{1}{2} \ln ∣ 1 + 2 \tan \frac{1}{2} x ∣ + c

\int_{0}^{\frac{π}{2}} \frac{dx}{1 + \cos x + 2 \sin x} = \frac{1}{2} {[\ln ∣ 1 + 2 \tan \frac{1}{2} x ∣]}_{0}^{\frac{π}{2}}

= \frac{\ln 3}{2} .

Commented by peter frank last updated on 23/Feb/22

thank you

Answered by MJS_new last updated on 23/Feb/22

\underset{0}{\int^{π / 2}} \frac{d x}{1 + \cos x + 2 \sin x} =

[t = \tan \frac{x}{2} \to d x = \frac{2 d t}{t^{2} + 1}]

= \underset{0}{\int^{1}} \frac{d t}{2 t + 1} = \frac{1}{2} {[\ln (2 t + 1)]}_{0}^{1} = \frac{\ln 3}{2}

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