Q 18 Evaluate ∫5 sinx2-cosx4-sin2xdx - Maths - Integrals

Question

Q 18 Evaluate ∫ 5   sin x 2 - cos x 4 - sin 2 x d x

Shruti Tyagi · Accepted Answer

Dear student,

Let I=∫5sinx2-cosx4-sin2xdx=5∫sinxcosx-2sin2x-4dx             --[taking -ve common from both the bracket of Dn]Put sin2x=1-cos2x=5∫sinxcosx-21-cos2x-4dx=5∫-sinxcosx-2cos2x+3dx         --[taking -ve common from IInd the bracket of Dn]Put u=cosxOn differentiatingdu=-sinx dxI=5∫1u-2u2+3du      ---(1)Using Partial fraction1u-2u2+3=Au-2+Bu2+3⇒1u-2u2+3=u2+3A+u-2Bu-2u2+3⇒1=u2+3A+u-2B        --[on cancelling the Dn]    ---(2)Put u=2⇒1=4+3A+0
B ⇒1=7A⇒A=17Put A=17 in eq (2)⇒1=u2+317+u-2B ⇒1-u2+317=u-2B ⇒(u-2)B=7-u2-37⇒B=4-u27(u-2)⇒B=-2+u2-u72-u      --[using a2-b2=a+ba-b]⇒B=-u+27⇒B=-u+27thus I=5∫17u-2-u+27u2+3duI=5∫17u-2du-5∫u+27u2+3du=57∫1u-2du-57∫u+2u2+3du=57logu-2-57∫uu2+3du+2u2+3       --[using ∫1xdx=logx]⇒I=57logu-2-57∫uu2+3du+107∫1u2+3du    ---(3)Let I1=∫uu2+3du  Put u2+3=vOn differentiating2udu=dvudu=dv2 I1=12∫1vdv=12log (v)undo substitutionI1=12logu2+3    ---[4]Let I2=∫1u2+3du=∫1u2+32du=13tan-1u3       --[5]    --[using ∫1x2+a2dx=1atan-1xa]Thus substituting the value of (4) and  (5) in (3) I=57logu-2-57
12logu2+3+107
13tan-1u3+C I=57logu-2-514logu2+3+1073tan-1u3+Cundo substitution u=cosx I=57logcosx-2-514logcos2x+3+1073tan-1cosx3+C

Regards

Q.18. Evaluate ∫ 5 sin x 2 - cos x 4 - sin 2 x d x .

Q.18. Evaluate $\int \frac{5 \sin x}{(2 - \cos x) (4 - \sin^{2} x)} d x$ .