Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.9 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.9

Integrate the following with respect to x:

Question 1.

e^{x} (tan x + log sec x)

Answer:

Let I = ∫e^{x} (tan x + log sec x) dx

Take f(x) = log sec x

f'(x) = \(\frac{1}{\sec x}\) sec x tan x

f'(x) = tan x

[∫e^{x} [f (x) + f (x)] dx = e^{x} f(x) + c]

∴ I = e^{x} log |sec x| + c

Question 2.

Answer:

[∫ e^{x} [f (x) + f (x)] dx = e^{x} f(x) + c]

Question 3.

e^{x} sec x (1 + tan x)

Answer:

Let I = \(\mathrm{I}=\int e^{x}(\sec x+\sec x \tan x) d x\)

Take f(x) = sec x

f ‘ (x) = sec x tan x

This is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\) = e^{x} f(x) + c

= e^{x} sec x + c

Question 4.

Answer:

Take f(x) = tan x

f'(x) = sec^{2} x

[∫ e^{x} [f (x) + f (x)] dx = e^{x} f(x) + c]

I = e^{x} tan x + c

Question 5.

Answer:

Put tan^{-1} x = t

\(\frac{1}{1+x^{2}}\)dx = dt

x = tan t

∴ I = ∫e^{t} [1 + tan t + tan^{2}t] dt

= ∫e^{t} [1 + tan^{2} t + tan t] dt

= ∫e^{t} [sec^{2}t + tan t] dt

= ∫e^{t} [tan t + sec^{2}t] dt

f(x) = tan t

f'(x) = sec^{2}t

[∫ e^{x} [f (x) + f (x)] dx = e^{x} f(x) + c]

∴ I = e^{t} tan t + c

I = e^{tan-1(x)} . x + c

I = x e^{tan-1(x)} + c

Question 6.

Answer: