Ex 7.3

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.3, 6 π ππ π₯ sinβ‘2π₯ sinβ‘3π₯ β«1βsinβ‘γπ₯ sinβ‘γ2π₯ sinβ‘3π₯ γ γ ππ₯ =β«1βγ(sinβ‘π₯ sinβ‘2π₯ ) sinβ‘3π₯ γ ππ₯ We know that 2 sinβ‘π΄ sinβ‘π΅=βcosβ‘(π΄+π΅)+cosβ‘(π΄βπ΅) sinβ‘π΄ sinβ‘π΅=1/2 [βcosβ‘(π΄+π΅)+cosβ‘(π΄βπ΅) ] sinβ‘π΄ sinβ‘π΅=1/2 [cosβ‘(π΄βπ΅)βcosβ‘(π΄+π΅) ] Replace A by π₯ & B by 2π₯ sinβ‘π₯ sinβ‘2π₯=1/2 [cosβ‘(π₯β2π₯)βcosβ‘(π₯+2π₯) ] sinβ‘π₯ sinβ‘2π₯ =1/2 [cosβ‘(βπ₯)βcosβ‘(3π₯) ] sinβ‘π₯ sinβ‘2π₯ =1/2 [cosβ‘γ π₯γβcosβ‘3π₯ ] Thus, our equation becomes β«1βπ¬π’π§β‘γπ π¬π’π§β‘ππ sinβ‘3π₯ γ ππ₯ =β«1βγπ/π (πππβ‘πβπππβ‘ππ ) γ . sinβ‘3π₯.ππ₯ =1/2 β«1β(cosβ‘π₯βcosβ‘3π₯ ) sinβ‘3π₯ ππ₯ =1/2 [β«1β(cosβ‘π₯. sinβ‘3π₯βcosβ‘3π₯. sinβ‘3π₯ ) ]ππ₯ =1/2 [β«1βγcosβ‘π₯. sinβ‘3π₯ γ ππ₯ββ«1βγcosβ‘3π₯. sinβ‘3π₯ γ ππ₯] (β΅πππ β‘(βπ₯)=πππ β‘π₯) β«1βγπππβ‘π. πππβ‘ππ γ ππ We know that 2 π ππβ‘π΄ πππ β‘π΅ =π ππβ‘(π΄+π΅)+π ππβ‘(π΄βπ΅) π ππβ‘π΄ πππ β‘π΅=1/2 [π ππβ‘(π΄+π΅)+π ππβ‘(π΄βπ΅) ] Replace A by 3π₯ & B by π₯ sinβ‘3π₯ cosβ‘π₯ = 1/2 [π ππβ‘(π₯+3π₯)+sinβ‘(3π₯βπ₯) ] = 1/2 [π ππβ‘4π₯+sinβ‘2π₯ ] β«1βγπππβ‘ππ. πππβ‘ππ γ ππ We know that 2 π ππβ‘π΄ πππ β‘π΅ =π ππβ‘(π΄+π΅)+π ππβ‘(π΄βπ΅) π ππβ‘π΄ πππ β‘π΅ =1/2 [π ππβ‘(π΄+π΅)+π ππβ‘(π΄βπ΅) ] Replace A by 3π₯ & B by 3π₯ sinβ‘3π₯ cosβ‘3π₯ = 1/2 [π ππβ‘(3π₯+3π₯)+sinβ‘(3π₯β3π₯) ] = 1/2 [π ππβ‘6π₯+sinβ‘0 ] =1/2 [π ππβ‘6π₯ ] Hence β«1βγsinβ‘3π₯.cosβ‘π₯ γ ππ₯ =1/2 β«1β[π ππβ‘4π₯+sinβ‘2π₯ ] ππ₯ Hence β«1βγcosβ‘3π₯.sinβ‘3π₯ γ ππ₯ =1/2 β«1βsinβ‘6π₯ ππ₯ Thus, our equation becomes β«1βsinβ‘γπ₯ sinβ‘γ2π₯ sinβ‘3π₯ γ γ ππ₯ =1/2 [β«1βγsinβ‘3π₯ cosβ‘3π₯ γ ππ₯ββ«1βγsinβ‘3π₯ cosβ‘3π₯ γ ππ₯] =1/2 [1/2 β«1β(sinβ‘4π₯+sinβ‘2π₯ ) ππ₯β1/2 β«1β(sinβ‘6π₯ ) ππ₯] =1/4 [β«1β(sinβ‘4π₯+sinβ‘2π₯ ) ππ₯ββ«1β(sinβ‘6π₯ ) ππ₯] =1/4 [β«1βsinβ‘4π₯ ππ₯+β«1βsinβ‘2π₯ ππ₯ββ«1βsinβ‘6π₯ ππ₯] β«1βsinβ‘(ππ₯+π) ππ₯=βπππ β‘(ππ₯ + π)/π +πΆ =1/4 [(βcosβ‘4π₯)/4 +(γβcosγβ‘2π₯/2) β((βcosβ‘6π₯)/6)]+πΆ =1/4 [(βcosβ‘4π₯)/4 βcosβ‘2π₯/2+cosβ‘6π₯/6]+πΆ =π/π [πππβ‘ππ/π βπππβ‘ππ/π β πππβ‘ππ/π ]+πͺ