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NIKITA PUBLICATION-INTEGRATION -MULTIPLE CHOICE QUESTIONS
- intsqrt(1+sinx)dx
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- intsqrt(1+sin2x)dx=
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- Evaluate int(1-cosx)"cosec"^(2)xdx
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- int(1)/(sin^(2)xcos^(2)x)dx=?
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- int(1)/(sin^(2)xcos^(2)x)dx=?
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- intsec^(2)xcosec^(2)x dx=
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- int(cos2x+2sin^(2)x)/(cos^(2)x)=
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- int(costheta-cos2theta)/(1-cos theta)d theta=
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- int(cos7x-cos8x)/(1+2cos 5x)dx=
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- int(cos2x)/(sin^(2)xcos^(2)x)dx=
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- int(cos4x)/(sin^(2)x)dx=
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- int(sin^(4)2x+cos^(4)2x)/(sin^(2)2xcos^(2)2x)dx=
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- int(sin^(6)theta+cos^(6)theta)/(sin^(2)theta cos^(2)theta)d theta=
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- int(cosec theta-cot theta)/(cosectheta+ cot theta)d theta=
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- int(cos2x-cos2alpha)/(cosx-cosalpha)dx
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- int(sinx+cosecx)/(tanx)dx=
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- 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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