Notes On Integrals of Some More Types - CBSE Class 12 Maths
The following integrals are solved in a particular way. âˆ« âˆš(x2 - a2) dx (or) âˆ« âˆš(a2 - x2) dx (or) âˆ« âˆš(x2 + a2) dx 1) âˆ« âˆš(x2 - a2) dx I = âˆ« âˆš(x2 - a2) dx I = âˆ« 1. âˆš(x2 - a2) dx I = âˆ« âˆš(x2 - a2) . 1dx Let f(x) = âˆš(x2 - a2) and g(x) = 1 âˆ« f(x) g(x) dx = f(x) âˆ« g(x) dx - âˆ« [ f '(x) âˆ« g(x) dx]  dx âˆ« âˆš(x2 - a2) . 1dx = âˆš(x2 - a2) . âˆ« 1 dx - âˆ« [d/dx(âˆš(x2 - a2)) âˆ«1dx] dx â‡’ I = âˆš(x2 - a2) . x - âˆ« 1/(2(âˆš(x2 - a2))) . 2x .x dx â‡’ I = xâˆš(x2 - a2) - âˆ« x2/âˆš(x2 - a2) dx â‡’ I = xâˆš(x2 - a2) - âˆ« (x2 - a2 + a2)/âˆš(x2 - a2) dx â‡’ I = xâˆš(x2 - a2) - âˆ« (x2 - a2)/âˆš(x2 - a2) dx - âˆ« a2/âˆš(x2 - a2) dx â‡’ I = xâˆš(x2 - a2) - âˆ« âˆš(x2 - a2) dx - âˆ« a2/âˆš(x2 - a2) dx â‡’ I = x âˆš(x2 - a2) - I - âˆ« a2/âˆš(x2 - a2) dx â‡’ 2I = x âˆš(x2 - a2) - a2 âˆ« 1/âˆš(x2 - a2) dx â‡’ 2I = x âˆš(x2 - a2) - a2 log|x + âˆš(x2 - a2)| + C â‡’ I = x/2 . âˆš(x2 - a2) - a2/2 . log|x + âˆš(x2 - a2)| + C Note: The integral of âˆš(x2 - a2) can also be evaluated by substituting x = a sec Î¸. 2) âˆ« âˆš(x2 + a2) dx I = âˆ« âˆš(x2 + a2) . 1dx Let  f(x) = âˆš(x2 + a2) and g(x) =1 âˆ« f(x) g(x) dx = f(x) âˆ« g(x) dx - âˆ« [ f '(x) âˆ« g(x) dx]  dx I = âˆš(x2 + a2) âˆ« 1 dx - âˆ« [d/dx (âˆš(x2 + a2)) âˆ« 1dx] dx â‡’ I = âˆš(x2 + a2) . x - âˆ« 1/(2âˆš(x2 + a2)). 2x . x dx â‡’ I = x âˆš(x2 + a2) - âˆ« x2/âˆš(x2 + a2) dx â‡’ I = x âˆš(x2 + a2) - âˆ« (x2 + a2 - a2)/âˆš(x2 + a2) dx â‡’ I = xâˆš(x2 + a2) - âˆ« (x2 - a2)/âˆš(x2 + a2) dx + âˆ« a2/âˆš(x2 + a2) dx â‡’ I = xâˆš(x2 + a2) - âˆ« âˆš(x2 + a2) dx + âˆ« a2/âˆš(x2 + a2) dx â‡’ I = x âˆš(x2 + a2) - I + âˆ« a2/âˆš(x2 + a2) dx â‡’ 2I = x âˆš(x2 + a2) + âˆ« a2 /âˆš(x2 + a2) dx â‡’ 2I = x âˆš(x2 + a2) +  a2 âˆ« 1 /âˆš(x2 + a2) dx â‡’ 2I = x âˆš(x2 - a2) + a2 log|x + âˆš(x2 + a2)| + C â‡’ I = x/2 . âˆš(x2 + a2) + a2/2 . log|x + âˆš(x2 + a2)| + C Note: The integral of âˆš(x2 + a2) can also be evaluated by substituting x = a tan Î¸. 3) âˆ« âˆš(a2 - x2) dx  = âˆ« âˆš(a2 - x2) . 1 dx      âˆ« âˆš(a2 - x2) . 1dx Let  f(x) = âˆš(a2 - x2) and g(x) = 1 âˆ« f(x) g(x) dx = f(x) âˆ« g(x) dx - âˆ« [ f '(x) âˆ« g(x) dx]  dx = âˆš(a2 - x2) âˆ« 1dx - âˆ« [d/dx (âˆš(a2 - x2)) âˆ« 1dx] dx â‡’ I = âˆš(a2 - x2) . x - âˆ« 1/2(âˆš(a2 - x2)) . (-2x) . x dx â‡’ I = xâˆš(a2 - x2) - âˆ« -x2/âˆš(a2 - x2) dx â‡’ I = xâˆš(a2 - x2) - âˆ« ((a2-x2)-a2)/âˆš(a2 - x2) dx â‡’ I = xâˆš(a2 - x2) - âˆ« (a2-x2)/âˆš(a2 - x2) dx + âˆ« a2/âˆš(a2 - x2) dx â‡’ I = xâˆš(a2 - x2) - âˆ« âˆš(a2 - x2) dx + âˆ« a2/âˆš(a2 - x2) dx â‡’ I = xâˆš(a2 - x2) - I + âˆ« a2/âˆš(a2 - x2) dx â‡’ 2I = xâˆš(a2 - x2) + a2 âˆ« 1/âˆš(a2 - x2) dx â‡’ 2I = xâˆš(a2 - x2) + a2 sin-1 (x/a) + C â‡’ I = x/2 âˆš(a2 - x2)+ a2/2 sin-1 (x/a) + C Note: The integral of âˆš(a2 - x2) can also be evaluated by substituting x = a sin Î¸.

#### Summary

The following integrals are solved in a particular way. âˆ« âˆš(x2 - a2) dx (or) âˆ« âˆš(a2 - x2) dx (or) âˆ« âˆš(x2 + a2) dx 1) âˆ« âˆš(x2 - a2) dx I = âˆ« âˆš(x2 - a2) dx I = âˆ« 1. âˆš(x2 - a2) dx I = âˆ« âˆš(x2 - a2) . 1dx Let f(x) = âˆš(x2 - a2) and g(x) = 1 âˆ« f(x) g(x) dx = f(x) âˆ« g(x) dx - âˆ« [ f '(x) âˆ« g(x) dx]  dx âˆ« âˆš(x2 - a2) . 1dx = âˆš(x2 - a2) . âˆ« 1 dx - âˆ« [d/dx(âˆš(x2 - a2)) âˆ«1dx] dx â‡’ I = âˆš(x2 - a2) . x - âˆ« 1/(2(âˆš(x2 - a2))) . 2x .x dx â‡’ I = xâˆš(x2 - a2) - âˆ« x2/âˆš(x2 - a2) dx â‡’ I = xâˆš(x2 - a2) - âˆ« (x2 - a2 + a2)/âˆš(x2 - a2) dx â‡’ I = xâˆš(x2 - a2) - âˆ« (x2 - a2)/âˆš(x2 - a2) dx - âˆ« a2/âˆš(x2 - a2) dx â‡’ I = xâˆš(x2 - a2) - âˆ« âˆš(x2 - a2) dx - âˆ« a2/âˆš(x2 - a2) dx â‡’ I = x âˆš(x2 - a2) - I - âˆ« a2/âˆš(x2 - a2) dx â‡’ 2I = x âˆš(x2 - a2) - a2 âˆ« 1/âˆš(x2 - a2) dx â‡’ 2I = x âˆš(x2 - a2) - a2 log|x + âˆš(x2 - a2)| + C â‡’ I = x/2 . âˆš(x2 - a2) - a2/2 . log|x + âˆš(x2 - a2)| + C Note: The integral of âˆš(x2 - a2) can also be evaluated by substituting x = a sec Î¸. 2) âˆ« âˆš(x2 + a2) dx I = âˆ« âˆš(x2 + a2) . 1dx Let  f(x) = âˆš(x2 + a2) and g(x) =1 âˆ« f(x) g(x) dx = f(x) âˆ« g(x) dx - âˆ« [ f '(x) âˆ« g(x) dx]  dx I = âˆš(x2 + a2) âˆ« 1 dx - âˆ« [d/dx (âˆš(x2 + a2)) âˆ« 1dx] dx â‡’ I = âˆš(x2 + a2) . x - âˆ« 1/(2âˆš(x2 + a2)). 2x . x dx â‡’ I = x âˆš(x2 + a2) - âˆ« x2/âˆš(x2 + a2) dx â‡’ I = x âˆš(x2 + a2) - âˆ« (x2 + a2 - a2)/âˆš(x2 + a2) dx â‡’ I = xâˆš(x2 + a2) - âˆ« (x2 - a2)/âˆš(x2 + a2) dx + âˆ« a2/âˆš(x2 + a2) dx â‡’ I = xâˆš(x2 + a2) - âˆ« âˆš(x2 + a2) dx + âˆ« a2/âˆš(x2 + a2) dx â‡’ I = x âˆš(x2 + a2) - I + âˆ« a2/âˆš(x2 + a2) dx â‡’ 2I = x âˆš(x2 + a2) + âˆ« a2 /âˆš(x2 + a2) dx â‡’ 2I = x âˆš(x2 + a2) +  a2 âˆ« 1 /âˆš(x2 + a2) dx â‡’ 2I = x âˆš(x2 - a2) + a2 log|x + âˆš(x2 + a2)| + C â‡’ I = x/2 . âˆš(x2 + a2) + a2/2 . log|x + âˆš(x2 + a2)| + C Note: The integral of âˆš(x2 + a2) can also be evaluated by substituting x = a tan Î¸. 3) âˆ« âˆš(a2 - x2) dx  = âˆ« âˆš(a2 - x2) . 1 dx      âˆ« âˆš(a2 - x2) . 1dx Let  f(x) = âˆš(a2 - x2) and g(x) = 1 âˆ« f(x) g(x) dx = f(x) âˆ« g(x) dx - âˆ« [ f '(x) âˆ« g(x) dx]  dx = âˆš(a2 - x2) âˆ« 1dx - âˆ« [d/dx (âˆš(a2 - x2)) âˆ« 1dx] dx â‡’ I = âˆš(a2 - x2) . x - âˆ« 1/2(âˆš(a2 - x2)) . (-2x) . x dx â‡’ I = xâˆš(a2 - x2) - âˆ« -x2/âˆš(a2 - x2) dx â‡’ I = xâˆš(a2 - x2) - âˆ« ((a2-x2)-a2)/âˆš(a2 - x2) dx â‡’ I = xâˆš(a2 - x2) - âˆ« (a2-x2)/âˆš(a2 - x2) dx + âˆ« a2/âˆš(a2 - x2) dx â‡’ I = xâˆš(a2 - x2) - âˆ« âˆš(a2 - x2) dx + âˆ« a2/âˆš(a2 - x2) dx â‡’ I = xâˆš(a2 - x2) - I + âˆ« a2/âˆš(a2 - x2) dx â‡’ 2I = xâˆš(a2 - x2) + a2 âˆ« 1/âˆš(a2 - x2) dx â‡’ 2I = xâˆš(a2 - x2) + a2 sin-1 (x/a) + C â‡’ I = x/2 âˆš(a2 - x2)+ a2/2 sin-1 (x/a) + C Note: The integral of âˆš(a2 - x2) can also be evaluated by substituting x = a sin Î¸.

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