Let's go back a couple years
∫ − π / 2 π / 2 2 0 0 7 x + 1 1 ⋅ sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x d x = ?
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2 solutions
When did you learn these? (I am a math enthusiast from class 10!)
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I learnt integration in class 12 CBSE :)
Where are you take such questions?
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I am not able to understand your question, please elaborate :)
I = ∫ − 2 π 2 π 2 0 0 7 x + 1 1 ⋅ sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x d x = ∫ − 2 π 0 2 0 0 7 x + 1 1 ⋅ sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x d x + ∫ 0 2 π 2 0 0 7 x + 1 1 ⋅ sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x d x = ∫ 0 2 π 2 0 0 7 − x + 1 1 ⋅ sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x d x + ∫ 0 2 π 2 0 0 7 x + 1 1 ⋅ sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x d x = ∫ 0 2 π 1 + 2 0 0 7 x 2 0 0 7 x ⋅ sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x d x + ∫ 0 2 π 2 0 0 7 x + 1 1 ⋅ sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x d x Replace x with − x
= ∫ 0 2 π sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x d x = 2 1 ∫ 0 2 π ( sin 2 0 0 8 x + cos 2 0 0 8 x sin 2 0 0 8 x + cos 2 0 0 8 x + sin 2 0 0 8 x cos 2 0 0 8 x ) d x = 2 1 ∫ 0 2 π d x = 4 π Using ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x