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Mathematics: Post your doubts here!

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No. It wouldve been easier with a substitution though.
But The reason is that d(arctan x)/dx = 1/(1+x^2)

and the integral is the opposite of the derivative

but how is it possible to solve that with substitution ? and which substitution will be used?
 
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but how is it possible to solve that with substitution ? and which substitution will be used?
If you applied the correct identity it might help . . . tan2(x)+1=sec2(x) Your integral equation then becomes, ∫1x2+1dx=∫dθ

^^This is what a teacher said. :confused:
 
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PLZ help me with part 2 !!
ii)Take 2 out.

2 ʃ ((2-u)^2 / u) du

Solve the square.

2 ʃ (4 - 4u +u^2)/u

Make fractions.

2 ʃ 4/u – 4u/u + u^2 /u)

Cancelling out will give you.

2 ʃ 4/u – 4 + u

Now integrate.

2 [ 4ln IuI – 4u + u^2/2 ]

8ln IuI - 8u + u^2

Plug in the limits.

(8ln I2I – 8(2) + (2^2)) – (8ln I1I – 8(1) + 1^2)

(8ln I2I – 16 + 4) – (0 – 8 + 1)

(8ln I2I – 12) – ( - 7)

8ln I2I - 12 + 7

8ln I2I – 5 Ans.
 
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ii)Take 2 out.

2 ʃ ((2-u)^2 / u) du

Solve the square.

2 ʃ (4 - 4u +u^2)/u

Make fractions.

2 ʃ 4/u – 4u/u + u^2 /u)

Cancelling out will give you.

2 ʃ 4/u – 4 + u

Now integrate.

2 [ 4ln IuI – 4u + u^2/2 ]

8ln IuI - 8u + u^2

Plug in the limits.

(8ln I2I – 8(2) + (2^2)) – (8ln I1I – 8(1) + 1^2)

(8ln I2I – 16 + 4) – (0 – 8 + 1)

(8ln I2I – 12) – ( - 7)

8ln I2I - 12 + 7

8ln I2I – 5 Ans.

Thanks a lot :)
 
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