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10 questions from this quiz
Integrals with lower indices like In−1I_{n-1}In−1
∫udvdxdx=uv−∫vdudxdx\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx∫udxdvdx=uv−∫vdxdudx
They avoid repeated manual integration
As the part containing power nnn
In=−nIn−1I_n = -nI_{n-1}In=−nIn−1
In−2I_{n-2}In−2
I0I_0I0
Yes, for both definite and indefinite
As sinx⋅sinn−1x\sin x \cdot \sin^{n-1} xsinx⋅sinn−1x
I0I_0I0 or I1I_1I1
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