(i) Integration by substitution or change of independent of independent variable.
(ii) Integration by parts.
(iii) Integration of rotational algebraic function by using partial fractions.
Integration by substitution: If φ (x) is a continuous differentiable function, then to evaluate integrals of the form ∫f (φ(x)) φ’ (x) dx,
We substitute φ (x) = t and φ’ (x) dx = dt
We substitution reduces the integral ∫f(φ(x)) φ’ (x) dx
This substitution reduces the integral ∫f(φ(x)) φ’ (x) dx to the form ∫f (t) dt
∫f (φ(x)) φ’ (x) dx = ∫f (φ(x)) d (φ (x)) [∵ dφ (x) = φ’ (x) dx]
= ∫f (t) dt, where t = φ(x)
Example: ex(x + 1) cos² (x.ex) dx
Solution: we observe that ex (x + 1) occurs in the derivative x.ex. So let us substitute x.ex
= t
⇒ d (x.ex).dx = dt
⇒ (x + 1) ex dx = dt
⇒
∴ ∫ ex(x + 1) cos² (x.ex) dx
= ∫cos²t dt
= ½ ∫ (1 + cos2t) dt
Algorithm to evaluate integrals of the form: ∫sinmx cosn x dx, ∫sinm x dx and ∫cosn x dx, where m, n ϵ N
Step I: find m and n.
Step II: if m is odd i.e., power or index of sin x is odd, put cosx = t and reduce the integral in terms of t.
If n is odd i.e., power of cosx is odd, then put sinx=t, and reduce the integral in terms of t.
If m and n both are odd, then either of the above substitutions can be used.
Step III: evaluate the integral obtained in step II and replace t by its value.
Example: ∫x² sin³ x³ cos⁵³ dx
Solution: let I = ∫x² sin³ x³ cos⁵³ dx
Here, indices of both sine and cosine are odd. So, we may substitute sinx³ = t or cos³ = t.
Let sinx³ = t then,
d (sinx³) = dt
⇒ cosx³. 3x² dx = dt
⇒
∴
= ⅓ ∫t³ cos⁴ x³ dt
= ⅓ ∫t³ (1 - sin² x³)² dt
= ⅓ ∫t³ (1 - t²)² dt
= ⅓ ∫(t³ - 2t⁵ + t⁷) dt
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