Example 6 - Chapter 7 Class 12 Integrals

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Example 6 Find the following integrals: (i) sin 3 cos 2 Let cos = Differentiate both sides . . . . sin = = sin Now are equation becomes sin 3 cos 2 Put the value of and = sin 3 . 2 . = sin 3 . 2 . sin = sin 3 sin 2 . = sin 2 2 . = 1 cos 2 2 . = 1 2 2 . = 2 4 = 2 + 4 = 2 . + 4 . = 2 +1 2 + 1 + C 1 + 4 + 1 4 + 1 + C 2 = 3 3 + 1+ 5 5 + 2 Putting back value of t = cos x = + + Example 6 Find the following integrals (ii) sin sin + Let + = Differentiate both sides . . . . 1= = Hence, our equation becomes sin sin + Putting the value of + and = sin sin = sin = sin = sin cos sin sin cos sin = sin cos sin . sin cos sin . = cos . sin co . = cos 1 . sin . = cos sin + 1 = . cos sin log sin + 1 Putting back value of t = x + a = + cos sin log sin + + 1 = cos + cos sin log sin + + 3 = cos sin log sin + +C Example 6 Find the following integrals (iii) 1 1 + tan 1 1 + tan The given function cannot be integrated by direct substitution, so we have to simplify the given function . Simplifying the given function and integrating. 1 1 + tan . = 1 1 + sin cos . = 1 1 + cos + sin cos . = cos sin + cos . Multiplying and dividing by 2 = 2 cos 2 sin + cos . = cos + cos 2 sin + cos . Adding and subtracting sin in the numerator = cos + cos + sin sin 2 sin + cos . = 1 2 cos + cos + sin sin sin + cos . = 1 2 sin + cos sin + cos + cos sin sin + cos = 1 2 1+ cos sin sin + cos = 1 2 1. + cos sin sin + cos = 1 2 + 1+ cos sin sin + cos Take, I1 = cos sin sin + cos . This can be solved by using the method of substitution Let sin + cos = Differentiate both sides . . . . cos sin = = cos sin Put the value of sin x + cos x and dx in I1 . I1= cos sin sin + cos . I1 = cos sin . cos sin I1 = 1 . I1 = log + 2 I1 = log sin + cos + 2 Put the value of I1 in eq. (1) 1 1 + tan = 1 2 + 1+ cos sin sin + cos = 1 2 + 1+ log sin + cos + 2 = 2 + 1 2 + 1 2 log sin + cos + 2 2 = 2 + 1 2 log sin + cos +