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Integration of ∫ (2-tanx)^1/2 . i have searched whole of google and also tried using the method of tanx^1/2 but reaching to no conclusion

 Integration of  ∫(2-tanx)^1/2 . i have searched whole of google and also tried using the method of tanx^1/2 but reaching to no conclusion

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
  
integral sqrt(2 - tan(x)) dx
For integrand sqrt(2 - tan(x)), substitute u= 2 - tan(x) and  du= -sec^2(x)  dx:
= integral -sqrt(u)/((2 - u)^2 + 1) du
=- integral sqrt(u)/((2 - u)^2 + 1) du
For integrand sqrt(u)/((2 - u)^2 + 1), substitute s= sqrt(u) and  ds= 1/(2 sqrt(u))  du:
=-2 integral s^2/((2 - s^2)^2 + 1) ds=-2 integral s^2/(s^4 - 4 s^2 + 5) ds
=-2 integral s^2/((s^2 - (2 + i)) (s^2 - (2 - i))) ds
using partial fractions:
=-2 integral ((1/2 + i)/(s^2 - (2 - i)) + (1/2 - i)/(s^2 - (2 + i))) ds
now integrating termwise-
=-1 - 2 i integral 1/(s^2 - (2 - i)) ds + -1 + 2 i integral 1/(s^2 - (2 + i)) ds
Factor -2 + i from denominator:
=-1 - 2 i integral -(2/5 + i/5)/(1 - (2/5 + i/5) s^2) ds + -1 + 2 i integral 1/(s^2 - (2 + i)) ds
=i integral 1/(1 - (2/5 + i/5) s^2) ds + -1 + 2 i integral 1/(s^2 - (2 + i)) ds
For integrand 1/(1 - (2/5 + i/5) s^2), substitute p= sqrt(-2/5 - i/5) s and  dp= sqrt(-2/5 - i/5)  ds:
=(-1 - 2 i) sqrt(-2/5 - i/5) integral 1/(p^2 + 1) dp + -1 + 2 i integral 1/(s^2 - (2 + i)) ds
 integral of 1/(p^2 + 1) is tan^(-1)(p):
=(-1 - 2 i) sqrt(-2/5 - i/5) tan^(-1)(p) + -1 + 2 i integral 1/(s^2 - (2 + i)) ds
Factor -2 - i from denominator:
=(-1 - 2 i) sqrt(-2/5 - i/5) tan^(-1)(p) - i integral 1/(1 - (2/5 - i/5) s^2) ds
For integrand 1/(1 - (2/5 - i/5) s^2), substitute w= sqrt(-2/5 + i/5) s and  dw= sqrt(-2/5 + i/5)  ds:
=(-1 - 2 i) sqrt(-2/5 - i/5) tan^(-1)(p) + (-1 + 2 i) sqrt(-2/5 + i/5) integral 1/(w^2 + 1) dw
 integral of 1/(w^2 + 1) is tan^(-1)(w):
=(-1 - 2 i) sqrt(-2/5 - i/5) tan^(-1)(p) - (1 - 2 i) sqrt(-2/5 + i/5) tan^(-1)(w) + c
Substitute back for w= sqrt(-2/5 + i/5) s:
=(-1 - 2 i) sqrt(-2/5 - i/5) tan^(-1)(p) - (1 - 2 i) sqrt(-2/5 + i/5) tan^(-1)(sqrt(-2/5 + i/5) s) + c
Substitute back for p= sqrt(-2/5 - i/5) s:
=(-1 - 2 i) sqrt(-2/5 - i/5) tan^(-1)(sqrt(-2/5 - i/5) s) - (1 - 2 i) sqrt(-2/5 + i/5) tan^(-1)(sqrt(-2/5 + i/5) s) + c
Substitute back for s= sqrt(u):
=(-1 - 2 i) sqrt(-2/5 - i/5) tan^(-1)(sqrt(-2/5 - i/5) sqrt(u)) - (1 - 2 i) sqrt(-2/5 + i/5) tan^(-1)(sqrt(-2/5 + i/5) sqrt(u)) + c
Substitute back for u= 2 - tan(x):
=(-1 - 2 i) sqrt(-2/5 - i/5) tan^(-1)(sqrt(-2/5 - i/5) sqrt(2 - tan(x))) - (1 - 2 i) sqrt(-2/5 + i/5) tan^(-1)(sqrt(-2/5 + i/5) sqrt(2 - tan(x))) + c
=-((1 + 2 i) sqrt(-2 - i) tan^(-1)(sqrt(-2/5 - i/5) sqrt(2 - tan(x))) + (1 - 2 i) sqrt(-2 + i) tan^(-1)(sqrt(-2/5 + i/5) sqrt(2 - tan(x))))/sqrt(5) + c
So ans
= i (sqrt(2 - i) tanh^(-1)(sqrt(2 - tan(x))/sqrt(2 - i)) - sqrt(2 + i) tanh^(-1)(sqrt(2 - tan(x))/sqrt(2 + i))) + c
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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