Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Chapter 7 Class 12 Integrals

Serial order wise

Last updated at June 13, 2023 by Teachoo

Misc 29 Evaluate the definite integral β«_0^(π/4)βγ(sinβ‘π₯ + cosβ‘π₯)/(9 + 16 sinβ‘2π₯ ) γ ππ₯ β«_0^(π/4)βγ (sinβ‘π₯ + cosβ‘π₯)/(9 +16 sinβ‘2π₯ ) γ ππ₯ Let sinβ‘π₯βcosβ‘π₯=π‘ Differentiating both sides π€.π.π‘.π₯ cosβ‘π₯+sinβ‘π₯=ππ‘/ππ₯ ππ₯=ππ‘/(sinβ‘π₯ + cosβ‘π₯ ) Now, sinβ‘π₯βcosβ‘π₯=π‘ Squaring both sides (sinβ‘π₯βcosβ‘π₯ )^2=π‘^2 sin^2β‘π₯+cos^2β‘π₯β2 sinβ‘π₯ cosβ‘π₯=π‘^2 1β2 sinβ‘π₯ cosβ‘π₯=π‘^2 1βsinβ‘2π₯=π‘^2 1βπ‘^2=sinβ‘2π₯ Putting the values of dx and sinβ‘2x, we get β«_0^(π/4)βγ (sinβ‘π₯ + cosβ‘π₯)/(9 +16 sianβ‘2π₯ ) γ ππ₯=β«1_(β1)^0βγ(sinβ‘π₯ + cosβ‘π₯)/(9 +16 sinβ‘2π₯ )Γππ‘/(sinβ‘π₯ + cosβ‘π₯ )γ =β«_(β1)^0βγ 1/(9 +16 (1 β π‘^2 ) ) γ. ππ‘ =β«_(β1)^0βγ 1/(9 +16 β 16π‘^2 ) γ. ππ‘ =β«_(β1)^0βγ 1/(25 β 16π‘^2 ) γ. ππ‘ =1/16 β«_(β1)^0βγ 1/(25/16 β π‘^2 ) γ. ππ‘ =1/16 β«_(β1)^0βγ 1/((5/4)^2 β π‘^2 ) γ. ππ‘ = γ1/16 [1/(2 . 5/4) γ logγβ‘γ |( 5/4 + π‘)/( 5/4 β π‘)|γ ]γ_(β1)^0 = γ1/4 [1/10 γ logγβ‘γ |( 5 + 4π‘)/( 5 β 4π‘)|γ ]γ_(β1)^0 = 1/40 γlogβ‘γ |( 5 + 4π‘)/( 5 β 4π‘)|γγ_(β1)^0 = 1/40 [logβ‘|( 5 + 4(0))/( 5 β 4(0) )|βlogβ‘|( 5 + 4(β1))/( 5 β 4(β1) )| ] = 1/40 [logβ‘|( 5 + 0)/( 5 β 0)|βlogβ‘|( 5 β 4)/( 5 + 4)| ] = 1/40 [logβ‘γ5/5γβlogβ‘γ1/9γ ] = 1/40 [γlog γβ‘1βlogβ‘(1/9) ] = 1/40 (logβ‘1+logβ‘γ(1/9)^(β1) γ ) = 1/40 (0+logβ‘γ9)γ = π/ππ πππ π