Mathematics - Calculus
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Evaluate the integral:
1. ∫cos^3 x sin^4 x dx
2. ∫sin^3 x dx
3. ∫cos^3 x/3 dx
4. ∫x/sqrt 9-x^2 dx
5. ∫1/sqrt 25-x^2 dx
6. ∫x sqrt16-4x^2 dx
7. ∫t/(1-t^2) ^3/2 dt
8. ∫sqrt 4x^2+9/x^4 dx
9. ∫1/x sqrt4x^2 +16 dx
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Solution Summary
Complete, neat and step-by-step solution evaluating the integrals are provided.
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(1) ∫ cos^3x.sin^4x dx = ∫ (sin^4x)(cos^2x)(cos x)dx =
∫ (sin ^4x)[1 - sin^2x]cos x.dx
Let sin x = t then cos x.dx = dt. ∴ Given Integral = ∫ t^4(1 - t^2)dt =
∫ t^4.dt - ∫ t^6.dt = t^5/5 - t^7/7 + C = sin^5x/5 - sin^7x/7 + C.
(2) ∫ sin^3x.dx = ∫ sin^2x.sin x.dx = ∫ [1 - cos^2x].sin x.dx =
∫ sin x.dx - ∫ cos^2x.sin x.dx
Now, ∫ sin x.dx = -cos x. Consider ∫ cos^2x.sin x.dx. Let cos x = t then
-sin x.dx = dt. Therefore, this integral becomes ∫ t^2(-dt) = -∫ t^2.dt =
-t^3/3 = -cos^3x/3.
Therefore, Given Integral = -cos ...
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