# Lecture 1 Part 3 Integrals Class 12 Maths

## Part - 3

Logic for this lecture:
Same logic we have used in Part 2 of this lecture!

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Questions Discussed in this lecture:
32. $\int _{}^{}\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha } \, dx = 2(\sin x+x\cos \alpha )+C$

33. $\int _{}^{}\sin ^{-1} (\cos x)\, dx = \frac{\pi x}{2} -\frac{x^{2} }{2} +C$

34. $\int _{}^{}\sin ^{2} (2x+5)\, dx = \frac{x}{2} -\frac{1}{8} \sin (4x+10)+C$

35. $\int _{}^{}\sin ^{3} (2x+1)\, dx = -\frac{1}{2} \cos (2x+1)+\frac{1}{6} \cos ^{3} (2x+1)+C$

36. $\int _{}^{}\sin ^{4} x\, dx = \frac{3x}{8} -\frac{1}{4} \sin 2x+\frac{1}{32} \sin 4x+C$

37. $\int \sin x.\sin 2x\, dx = -\frac{1}{2} \left(\frac{\sin 3x}{3} -\sin x\right)+C$

38. $\int _{}^{}\sin 3x\cos 4x \, dx = -\frac{1}{14} \cos 7x+\frac{1}{2} \cos x+C$

39. $\int _{}^{}\tan ^{2} (2x-3)\, dx = \frac{1}{2} \tan (2x-3)-x+C$

40. $\int \sin ^{2} x\, \, \cos ^{4} x \, dx = \frac{1}{32} \left[2x+\frac{1}{2} \sin 2x-\frac{1}{2} \sin 4x-\frac{1}{6} \sin 6x\right]+C$

41. $\int _{}^{}\cos 2x\cos 4x\cos 6x \, dx = \frac{1}{4} \left[\frac{1}{12} \sin 12x+x+\frac{1}{8} \sin 8x+\frac{1}{4} \sin 4x\right]+C$

43. $\int _{}^{}\sin x\sin 2x\sin 3x\, dx = \frac{1}{4} \left[\frac{1}{6} \cos 6x-\frac{1}{4} \cos 4x-\frac{1}{2} \cos 2x\right]+C$

44. $\int \tan ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x} } \, dx = \frac{\pi }{4} x-\frac{x^{2} }{4} +C$

45. $\int _{}^{}\frac{e^{5\log x} -e^{4\log x} }{e^{3\log x} -e^{2\log x} } \, dx = \frac{x^{3} }{3} +C$

46. $\int _{}^{}\frac{dx}{\sqrt{1-2x} +\sqrt{3-2x} } = \frac{1}{6} (1- x)^{\frac{3}{2} } -\frac{1}{6} (3-2x)^{\frac{3}{2} } +C$

47. $\int \tan ^{-1} (\cot x)\, dx = \frac{\pi }{2} x-\frac{x^{2} }{2} +C$

48. $\int _{0}^{\pi }\, ( \sin ^{2} \frac{x}{2} -\cos ^{2} \frac{x}{2} )\, dx = 0$

49. $\int _{0}^{1}\frac{dx}{\sqrt{1+x} -\sqrt{x} } = \frac{4\sqrt{2} }{3}$