• Wed. Jul 24th, 2024
    • '
    Electrician

    Methods of Integration |Integration using Trigonometric Identities

    author
    3 minutes, 16 seconds Read

    Methods of Integration |Integration using Trigonometric Identities

    Integration by Substitution

    If we have to evaluate an integral of the type

    ∫f{Φ(x) Φ'(x)dx then we put Φ(x) =t and Φ'(x) dx = dt. With this substitution, the integrand becomes easily integrable.

    Case1 When the integrand is of the form
    f(ax + b), we put (ax +b)= t and dx =(1/a)dt

    Case II When the integrand is of the form xⁿ⁻¹.f(xⁿ), we put xⁿ = t and nxⁿ⁻¹dx = dt.

    Case III When the integrand is of the form
    {f(x)}ⁿ f'(x), we put f(x) = t and f'(x)dx=dt.

    Case IV When the integrand is of the form
    f'(x)/f(x) we put f(x) = t and f'(x)dx=dt.

    Theorem

    ∫(ax+b)ⁿdx={(ax+b)ⁿ⁺¹/a(n+1)}+C where n≠-1

    Proof
    Putting ax+b=t
    we get adx =dt or dx =(1/a)dt
    ∴ ∫(ax+b)ⁿdx=1/a∫ tⁿdt
    ={tⁿ⁺¹/a(n+1)}+C

    ∫(ax+b)ⁿdx =(ax+b)ⁿ⁺¹/a(n+1) +C

    Theorem

    1. ∫cos(ax+b)dx=(1/a)sin(ax+b)+C

    Proof

    Put(ax+b)=t so that dx=(1/a)dt ∴ ∫cos(ax+b)dx=(1/a)∫cost dt
              =(1/a)sint+C
            =(1/a).sin(ax+b)+C

    ∫cos(ax+b)dx=(1/a) sin(ax+b)+C

    (ii)∫cosec²(ax+b)dx = -(1/a) cot(ax+b)+C

    Proof

    Put(ax+b)=t so that dx=(1/a) dt
    cosec²(ax+b)dx= (1/a)∫cosec²t dt
    =-(1/a)cot t +C
    =-(1/a).cot(ax+b) +C

    For e.g

    (i) ∫tanxdx

    solution

    = ∫tanxdx
    =∫(sinx/cosx) dx
    =∫(1/t)dt [where cosx= t and sinxdx=-dt]
    = -log|t|+C
    ∫tanxdx =-log|cosx|+C

    ∴ ∫tanxdx =-log|cosx|+C

    (ii) ∫cotxdx

    solution

    ∫cotxdx=∫(cosx/sinx)dx
    =∫(1/t)dt [where sinx= t and cosxdx=-dt]
    = log|t|+C
    =log|sinx|+C

    ∴ ∫cotxdx =log|sinx|+C

    (iii)∫secxdx

    solution

    ∫secxdx=∫{secx(secx+tanx)/(secx+tanx)}dx

    [multiplying numerator and denominator by (secx+tanx)]

    =∫(1/t)dt

    [ where (secx+tanx)= t and secx(secx+tanx)dx=dt ]

    = log|t|+C
    =log| (secx+tanx) |+C

    ∴ ∫secxdx =log| (secx+tanx) |+C

    (iv) ∫cosecxdx

    solution

    ∫cosecxdx

    =∫{cosecx(cosecx-cotx)/(cosecx-cotx)}dx

    [multiplying numerator and denominator by (cosecx-cotx)]

    =∫(1/t)dt

    [ where (cosecx-cotx) = t and cosecx(cosecx-cotx)dx=dt ]

    = log|t|+C
    =log|(cosecx-cotx) |+C

    ∴ ∫cosecxdx =log|(cosecx-cotx) |+C

    Integration Using Trigonometric Identities

    When the integrand consists of trigonometric functions, we use known identities to convert it into a form which can easily be integrated. Some of the identities useful for this purpose are given below:

    (i)2sin²(x/2)=(1-cosx)

    (ii)2cos²(x/2)=(1+cosx)

    (iii)2sinacosb=sin(a+b)+sin(a-b)

    (iv)2cosasinb =sin(a+b)-sin(a-b)

    (v)2cosacosb=cos(a+b)+cos(a-b)

    (vi)2sinasinb=cos(a-b)-cos(a+b)

    For Example

    (i) ∫(sin3xsin2x)dx

    Solution

    ∫(sin3xsin2x)dx
    using 2sina sinb=cos(a-b)-cos(a+b) we have
    ∫(sin3xsin2x)dx=1/2∫(2sin3xsin2x)dx
    =1/2∫(cosx-cos5x)dx
    =1/2[∫(cosx)-∫(cos5x)dx
    ∫(sin3xsin2x)dx = (1/2)sinx – sin5x/10 +C

    (ii) ∫cos3xsin2xdx

      Solution

∫cos3xsin2xdx
using 2cosa sinb=sin(a+b)-sin(a-b)

    ∫cos3xsin2xdx =1/2∫(2cos3xsin2x)dx
    =1/2∫(sin5x-sinx)dx
    =1/2[∫(sin5x)-∫(sinx)dx]
    ∫cos3xsin2xdx =cos5x/10+cosx/2+C

    ∫cos3xsin2xdx=cos5x/10+cosx/2+C

    (iii) ∫(sin³xcos³x)dx

    solution

    ∫(sin³xcos³x)dx=∫(sin³xcos²xcosx)dx
    =∫(sin³x(1-sin²x)cosx)dx
    =∫t³(1-t²)dt [where sinx=t]
    =∫t³dt-∫t⁵dt
    =t⁴/4-t⁶/6+C
    ∫(sin³xcos³x)dx =sin⁴x/4-sin⁶x/6+C

    ∫(sin³xcos³x)dx=sin⁴x/4-sin⁶x/6+C

    (iv) ∫sin⁴xdx

    solution

    ∫sin⁴xdx=1/4∫(2sin²x)²dx
    =1/4∫(1-cos2x)²dx
    =1/4∫(1+cos²2x-2cos2x)dx
    =1/8∫(2+2cos²2x-4cos2x)dx
    =1/8∫[2+(1+cos4x)-4cos2x]dx
    =3/8∫dx+1/8∫(cos4x)dx-1/2∫(cos2x)dx
    =3/8x+sin4x/32-sin2x/4+C

    ∫sin⁴xdx =3/8x+sin4x/32-sin2x/4+C

    ये भी पढ़े….

    Similar Posts

    Leave a Reply