


Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 7.10
Ex 7.10, 2
Ex 7.10, 3 Important
Ex 7.10, 4
Ex 7.10, 5 Important
Ex 7.10, 6
Ex 7.10,7 Important
Ex 7.10,8 Important
Ex 7.10, 9
Ex 7.10, 10 Important
Ex 7.10, 11 Important
Ex 7.10, 12 Important You are here
Ex 7.10, 13
Ex 7.10, 14
Ex 7.10, 15
Ex 7.10, 16 Important
Ex 7.10, 17
Ex 7.10, 18 Important
Ex 7.10, 19
Ex 7.10, 20 (MCQ) Important
Ex 7.10, 21 (MCQ) Important
Last updated at May 29, 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₯ Using P4 : β«_0^πβγπ(π₯)ππ₯=γ β«_0^πβπ(πβπ₯)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