class 11 mathematics Question

Methods of Integration |Integration using Trigonometric Identities

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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

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