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Definite Integration by properties - P4
Ex 7.10, 2
Ex 7.10, 3 Important
Example 32 Important
Ex 7.10, 4
Ex 7.10, 15
Ex 7.10, 17
Ex 7.10, 21 (MCQ) Important
Ex 7.10, 19
Ex 7.10, 12 Important You are here
Ex 7.10, 9
Ex 7.10,7 Important
Example 30
Question 4 (MCQ) Important Deleted for CBSE Board 2024 Exams
Ex 7.10, 10 Important
Ex 7.10,8 Important
Example 34 Important
Ex 7.10, 16 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Example 42 Important
Definite Integration by properties - P4
Last updated at June 13, 2023 by Teachoo
Ex 7.10, 12 By using the properties of definite integrals, evaluate the integrals: β«_0^πβ(π₯ ππ₯)/(1 + sinβ‘π₯ ) ππ₯ Let I=β«_0^πβπ₯/(1+ sinβ‘π₯ ) ππ₯ β΄ I=β«_0^πβ(π β π₯)/(1+ sinβ‘π₯ ) ππ₯ Adding (1) and (2) i.e. (1) + (2) I+I=β«_0^πβ( π₯)/(1 + sinβ‘π₯ ) ππ₯+β«_0^πβ( π β π₯)/(1 + sinβ‘π₯ ) ππ₯ 2I=β«_0^πβ( π₯ + π β π₯)/(1 + sinβ‘π₯ ) ππ₯ 2I=β«_0^πβ( π)/(1 + sinβ‘π₯ ) ππ₯ I=π/2 β«_0^πβ( 1)/(1 + sinβ‘π₯ ) ππ₯ Multiplying and dividing by (1βsinβ‘π₯ ) I=π/2 β«_0^πβγ( 1)/(1 + sinβ‘π₯ ) Γ (1 β sinβ‘π₯)/(1 β sinβ‘π₯ )γ . ππ₯ I=π/2 β«_0^πβ(1 β sinβ‘π₯)/(1 β sin^2β‘π₯ ) ππ₯ I=π/2 β«_0^πβ(1 β sinβ‘π₯)/( γcos γ^2β‘π₯ ) ππ₯ I=π/2 β«_0^πβ[1/cos^2β‘π₯ βsinβ‘π₯/( γcos γ^2β‘π₯ )] ππ₯ I=π/2 β«_0^πβ[sec^2β‘π₯βsinβ‘π₯/(cosβ‘π₯ .γ cosγβ‘π₯ )] ππ₯ I=π/2 β«_0^πβ[sec^2β‘π₯βtanβ‘γπ₯ secβ‘π₯ γ ] ππ₯ I=π/2 [[tanβ‘π₯ ]_0^πβ[secβ‘π₯ ]_0^π ] I=π/2 [[π‘ππ(π)βπ‘ππ(0)]β[π ππ(π)βπ ππ(0)]] I=π/2 [[0β0]β[β1β1]] I=π/2 [0β(β2)] I=π/2 [2] π=π