Part - 3 Lecture - 1 Chapter 7 Integrals
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यही जज्बा रहा तो मुश्किलों का हल भी निकलेगा
जमीं बंजर हुई तो क्या वहीं से जल भी निकलेगा
ना हो मायूस ना घबरा अंधेरों से मेरे साथी
इन्हीं रातों के दामन से सुनहरा कल भी निकलेगा
Logic for this lecture:
There are a lot of ways to perform substitution in integration. First way is to substitute the function only if its derivative is also occurring in question. Sometimes, you need to simplify the function to know whether you can apply substitution or not.
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
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^{5log x} -e^{4log x} }{e^{3log x} -e^{2log 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}
