Ex 7.3, 10 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Ex 7.3
Ex 7.3, 1
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Transcript
Ex 7.3, 10 Integrate the function π ππ4 π₯ β«1βsin^4β‘π₯ ππ₯ =β«1β(sin^2β‘π₯ )^2 ππ₯ =β«1β((1 β cosβ‘2π₯)/2)^2 ππ₯ =1/4 β«1β(1βcosβ‘2π₯ )^2 ππ₯ We know that πππ β‘2π=1β2 γπ ππγ^2β‘π 2 γπ ππγ^2β‘π=1βπππ β‘2π γπ ππγ^2β‘π=(1 β πππ β‘2π)/2 Replace π by π₯ γπ ππγ^2β‘π₯=(1 β πππ β‘2π₯)/2 =1/4 β«1β(1^2+(cosβ‘2π₯ )^2β2(1)(cosβ‘2π₯ )) ππ₯ =1/4 β«1β(1+cos^2β‘2π₯β2 cosβ‘2π₯ ) ππ₯ =1/4 β«1β(1+(1 + cosβ‘4π₯)/2β2 cosβ‘2π₯ ) ππ₯ =1/4 β«1β1 ππ₯+1/8 β«1β(1+cosβ‘4π₯ ) ππ₯β2/4 β«1βcosβ‘2π₯ ππ₯ =1/4 β«1β1 ππ₯+1/8 β«1β1 ππ₯+1/8 β«1βcosβ‘4π₯ ππ₯β1/2 β«1βcosβ‘2π₯ ππ₯ We know that cosβ‘2π=2 cos^2β‘γπβ1γ cosβ‘γ2π+1γ=2 cos^2β‘π Replace π by 2π₯ cosβ‘γ4π₯+1γ=2 cos^2β‘2π₯ cosβ‘γ4π₯ + 1γ/2=cos^2β‘2π₯ β«1βcosβ‘(ππ₯+π) ππ₯=π ππβ‘(ππ₯ + π)/π =π₯/4 + π₯/8 + 1/8 sinβ‘4π₯/4 β 1/2 sinβ‘2π₯/2 +πΆ =(2π₯ + π₯)/8 + 1/32 sinβ‘4π₯β 1/4 sinβ‘2π₯+πΆ =ππ/πβ π/π πππβ‘ππ++ π/ππ πππβ‘ππ+πͺ
