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options 0 -TT/2 7TT/2 TT/2

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Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

I = \int_{-1/2}^{1/2}[sin^{-1}(3x-4x^3)-cos^{-1}(4x^3-3x)]dx
I = \int_{-1/2}^{1/2}[sin^{-1}(-(4x^3-3x))-cos^{-1}(4x^3-3x)]dx
sin^{-1}(-t) = -sin^{-1}(t)
I = \int_{-1/2}^{1/2}[-sin^{-1}(4x^3-3x)-cos^{-1}(4x^3-3x)]dx
I = \int_{-1/2}^{1/2}-[sin^{-1}(4x^3-3x)+cos^{-1}(4x^3-3x)]dx
sin^{-1}t + cos^{-1}t = \frac{\pi }{2}
I = \int_{-1/2}^{1/2}-\frac{\pi }{2}dx
I = -\frac{\pi }{2}(x)_{-1/2}^{1/2}
I = -\frac{\pi }{2}

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