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NIKITA PUBLICATION-INTEGRATION -MULTIPLE CHOICE QUESTIONS
- int(sectheta)/(sec theta+tantheta)d theta=
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- If int(cos4x+1)/(cotx-tanx)dx=kcos4x+c, then k= (A) -1/4 (B) -1/2 (C) ...
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- int(1)/(tanx+cotx)dx=
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- int((1-tanx)/(1+tanx))^(2)dx=
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- int(dx)/((1-cosx))=?
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- int(dx)/(1+cosx)
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- Evaluate: (i) int1/(1+cos2x)\ dx (ii) int1/(1-cos2x)\ dx
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- int(1)/(1-sinx) dx
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- int(cosx)/(1-cosx)dx=
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- int(cotx)/(cosecx-cotx)dx=
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- int(sinx)/((1+sinx))dx=?
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- int(tanx)/(secx+tanx)dx=
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- int(cosx)/(sqrt(1+sinx))dx is equal to
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- int(sinx+cosx)/(sqrt(1+sinx))dx
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- Evaluate the following integrals : inttan^(-1)((sin2x)/(1+cos2x))dx
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- intcot^(-1)((1+sinx)/(cosx))dx=
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- Evaluate: inttan^(-1){sqrt(((1-sinx)/(1+sinx)))}\ dx ,\ -pi//2<x<pi//2
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- inttan^(-1)((2tanx)/(1-tan^(2)x))dx=
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- inttan^(-1)(sec3x+tan3x)dx=
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- intcot^(-1)sqrt((1+cosx)/(1-cosx))dx=
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