If I = ∫ − 2 π 2 π sin − 1 ( sin x ) d x then − π 2 16 I =

I = ∫ − 2 π 2 π x d x + ∫ 2 π 2 3 π ( π − x ) d x + ∫ 2 3 π 2 π ( x − 2 π ) d x = 8 − π 2

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationProperties of definite integrationMedium2 minQB

MathematicsMediuminteger

If $I = \int_{- \frac{π}{2}}^{2 π} sin^{- 1} (sin x) d x$ then $- \frac{16}{π ^{2}} I =$

Answer:

Solution:

$I = \int_{- \frac{π}{2}}^{\frac{π}{2}} x d x + \int_{\frac{π}{2}}^{\frac{3 π}{2}} (π - x) d x + \int_{\frac{3 π}{2}}^{2 π} (x - 2 π) d x = \frac{- π ^{2}}{8}$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Properties of definite integration

⏱ 2mℹ️ Source: QB

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