Lecture 1 Part 5 Integrals Class 12 Maths

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Part - 5 Lecture - 1 Chapter 7 Integrals

Anybody can become angry – that is easy, but to be angry with the right person and to the right degree and at the right time and for the right purpose, and in the right way – that is not within everybody’s power and is not easy.

Booklets/Notes/Assignments are typed here on this website and their PDFs will be made available soon.

Logic for this lecture:
In case trigonometric functions are given as a fraction you can use following identities in numerator to simplify it:
\(\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B \)
\(\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B\)

Questions Discussed in this lecture:
53. \(\int \frac{\sin (x-\alpha )}{\sin (x+\alpha )}  , dx=x\cos 2\alpha -\sin 2\alpha \log |\sin (x+\alpha )|+C\)

54. \(\int \frac{\cos (x+a)}{\cos (x-a)} , dx =x\cos 2a-\sin 2a, \log |\sec (x-a)|+C\)

55. \(\int \frac{\sin x}{\sin (x+a)} dx  = x\cos a-\sin a.\log |\sin (x+a)|+C \)

56. \(\int \frac{1}{\sin (x-a)\sin (x-b)} dx   = \frac{1}{\sin (a-b)} \log \left|\frac{\sin (x-a)}{\sin (x-b)} \right|+C\)

57. \(\int \frac{\sin x}{\sin (x-a)} , dx  =  \sin a\log |\sin (x-a)|+x\cos a+C \)

58. \(\int \frac{1}{\cos (x+a)\cos (x+b)}  , dx   =  \frac{1}{\sin (a-b)} \log \left|\frac{\cos (x+b)}{\cos (x+a)} \right|+C\)

59. \(\int \frac{\sin (x+a)}{\sin (x+b)} , dx   =  x\cos (a-b)+\sin (a-b)\log |\sin (x+b)|+C\)

60. \(\int \frac{\cos (x+a)}{\sin (x+b)} , dx  = \cos (a-b), \log |\sin (x+b)|-\sin (a-b).x+C\)

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