We will solve the above problem of integration by using substitution method :-
⟹
Now , we get :-
where c = constant
Now put t = √x
∫ 1 dx = x + C
∫ sin x dx = – cos x + C
∫ cos x dx = sin x + C
∫ sec2 dx = tan x + C
∫ csc2 dx = -cot x + C
∫ sec x (tan x) dx = sec x + C
∫ csc x ( cot x) dx = – csc x + C
∫ (1/x) dx = ln |x| + C
∫ ex dx = ex+ C
∫ ax dx = (ax/ln a) + C
GIVEN :-
TO FIND :-
SOLUTION :-
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We will solve the above problem of integration by using substitution method :-
⟹![let \: \sqrt{x} = t let \: \sqrt{x} = t](https://tex.z-dn.net/?f=let%20%5C%3A%20%20%5Csqrt%7Bx%7D%20%20%3D%20t)
⟹![(x = {t}^{2} ) (x = {t}^{2} )](https://tex.z-dn.net/?f=%28x%20%3D%20%20%7Bt%7D%5E%7B2%7D%20%29)
⟹![{x}^{ \frac{1}{2} } = t {x}^{ \frac{1}{2} } = t](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%20%20%3D%20t)
⟹![\dfrac{1}{2} {x}^{ \frac{ - 1}{2} } dx = dt \dfrac{1}{2} {x}^{ \frac{ - 1}{2} } dx = dt](https://tex.z-dn.net/?f=%20%5Cdfrac%7B1%7D%7B2%7D%20%20%7Bx%7D%5E%7B%20%5Cfrac%7B%20-%201%7D%7B2%7D%20%7D%20dx%20%3D%20dt)
⟹![\dfrac{dx}{2 \sqrt{x} } = dt \dfrac{dx}{2 \sqrt{x} } = dt](https://tex.z-dn.net/?f=%20%5Cdfrac%7Bdx%7D%7B2%20%5Csqrt%7Bx%7D%20%7D%20%20%3D%20dt)
⟹![dx = 2 \sqrt{x} dt dx = 2 \sqrt{x} dt](https://tex.z-dn.net/?f=dx%20%3D%202%20%5Csqrt%7Bx%7D%20dt)
⟹![dx = 2 t \: dt dx = 2 t \: dt](https://tex.z-dn.net/?f=dx%20%3D%202%20t%20%5C%3A%20dt)
Now , we get :-
⟹![\displaystyle \int \dfrac{2 {t}^{2} \: dt}{1 + {t}^{2} }\:dx \displaystyle \int \dfrac{2 {t}^{2} \: dt}{1 + {t}^{2} }\:dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B2%20%7Bt%7D%5E%7B2%7D%20%20%5C%3A%20dt%7D%7B1%20%2B%20%20%7Bt%7D%5E%7B2%7D%20%7D%5C%3Adx)
⟹![2\displaystyle \int \dfrac{ {t}^{2} + 1 - 1\: }{1 + {t}^{2} }\:dt 2\displaystyle \int \dfrac{ {t}^{2} + 1 - 1\: }{1 + {t}^{2} }\:dt](https://tex.z-dn.net/?f=2%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B%20%7Bt%7D%5E%7B2%7D%20%20%20%2B%201%20-%201%5C%3A%20%7D%7B1%20%2B%20%20%7Bt%7D%5E%7B2%7D%20%7D%5C%3Adt)
⟹![2\displaystyle \int \dfrac{ {t}^{2} + 1 \: }{1 + {t}^{2} }\:dt - 2\displaystyle \int \dfrac{ 1\: }{1 + {t}^{2} }\:dt 2\displaystyle \int \dfrac{ {t}^{2} + 1 \: }{1 + {t}^{2} }\:dt - 2\displaystyle \int \dfrac{ 1\: }{1 + {t}^{2} }\:dt](https://tex.z-dn.net/?f=2%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B%20%7Bt%7D%5E%7B2%7D%20%20%20%2B%201%20%5C%3A%20%7D%7B1%20%2B%20%20%7Bt%7D%5E%7B2%7D%20%7D%5C%3Adt%20-%202%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B%201%5C%3A%20%7D%7B1%20%2B%20%20%7Bt%7D%5E%7B2%7D%20%7D%5C%3Adt)
⟹![2\displaystyle \int1 \:dt - 2\displaystyle \int \dfrac{ 1\: }{1 + {t}^{2} }\:dt 2\displaystyle \int1 \:dt - 2\displaystyle \int \dfrac{ 1\: }{1 + {t}^{2} }\:dt](https://tex.z-dn.net/?f=2%5Cdisplaystyle%20%5Cint1%20%5C%3Adt%20-%202%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B%201%5C%3A%20%7D%7B1%20%2B%20%20%7Bt%7D%5E%7B2%7D%20%7D%5C%3Adt)
⟹![2t - 2 {\tan}^{ - 1} (t) + c 2t - 2 {\tan}^{ - 1} (t) + c](https://tex.z-dn.net/?f=2t%20-%202%20%20%7B%5Ctan%7D%5E%7B%20-%201%7D%20%28t%29%20%20%2B%20c)
where c = constant
Now put t = √x
_________________________
Learn more :-
∫ 1 dx = x + C
∫ sin x dx = – cos x + C
∫ cos x dx = sin x + C
∫ sec2 dx = tan x + C
∫ csc2 dx = -cot x + C
∫ sec x (tan x) dx = sec x + C
∫ csc x ( cot x) dx = – csc x + C
∫ (1/x) dx = ln |x| + C
∫ ex dx = ex+ C
∫ ax dx = (ax/ln a) + C
______________________
GIVEN :-
TO FIND :-
SOLUTION :-