Class 12 Integrals · Very Long Answer5 Marksdefinite_integration

Class 12 Maths · Integrals · Very Long Answer #118

For \(-\pi/2<\alpha<\pi/2\), evaluate \(\int_0^\pi x/(1+\sin\alpha\sin x)dx\).

✅ Step-by-Step Solution

Final Answer: π(π−2α)/(2cosα)

The denominator is positive for this α range. Reflection x↦π−x gives I=(π/2)∫₀^πdx/(1+sinα sin x).
Set u=tan(x/2), sin x=2u/(1+u²), dx=2du/(1+u²), with u from 0 to ∞. Then I=π∫₀^∞du/(u²+2u sinα+1).
Complete the denominator as (u+sinα)²+cos²α. Since cosα>0, an antiderivative is (1/cosα)arctan[(u+sinα)/cosα].
The limit at infinity is π/2 and the lower value is arctan(tanα)=α on the stated range. Thus I=π(π/2−α)/cosα.
Therefore the integral equals π(π−2α)/(2cosα).

[1 mark] The denominator is positive for this α range. Reflection x↦π−x gives I=(π/2)∫₀^πdx/(1+sinα sin x).
[1 mark] Set u=tan(x/2), sin x=2u/(1+u²), dx=2du/(1+u²), with u from 0 to ∞. Then I=π∫₀^∞du/(u²+2u sinα+1).
[1 mark] Complete the denominator as (u+sinα)²+cos²α. Since cosα>0, an antiderivative is (1/cosα)arctan[(u+sinα)/cosα].
[1 mark] The limit at infinity is π/2 and the lower value is arctan(tanα)=α on the stated range. Thus I=π(π/2−α)/cosα.
[1 mark] Therefore the integral equals π(π−2α)/(2cosα).

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