Integrals NCERT Solutions
NCERT Exercise 7.6 Q: 16 to 24
Question: 16. `e^x(sin x + cos x)`
Solution: `int e^x(sin x + cos x)dx`
`=int e^x sin x dx + int e^x cos x dx`
`=I_1 + int e^x cos x dx`
Now, `I_1 = int e^x sin x dx`
Taking ` sin x` as first function and ` e^x` as second function then
`:. I_1 = sin x int e^x dx -` `int[d/dx (sin x) * int e^x dx] dx`
`=sin x * e^x - int cos x * e^x dx `
`= e^x sin x - int e^x cos x + C`
Putting the value of `I_1` in (i), we get
` int e^x (sin x + cos x) dx`
`=e^x sin x - int e^x cos x dx + ` ` int e^x cos x dx + C`
`= e^x sin x + C`
Question: 17. `(x e^x)/(1+x)^x`
Solution: `int (x e^x)/(1+x)^x dx`
`=int e^x ((x+1-1)/(x+1)^2)dx`
`=int e^x (1/(x+1)-1/(x+1)^2)dx`
`=int e^x 1/(x+1) dx - int e^x * 1/(x+1)^2 dx`
`=I_1 - int e^x * 1/(x+1)^2 dx` ----(i)
Now, `I_1 = int e^x 1/(x+1) dx`
Taking `1/(x+1)` as first function `e^x` as second function then
`:. I_1 = 1/(x+1) int e^x dx -` `int[d/dx(1/(x+1)) * int e^x dx] dx`
`= 1/(x+1) e^x - int - 1/(x+1)^2 * e^x dx +C`
`= e^x/(x+1) + int e^x/(x+1)^2 dx + C`
Putting value of `I_1` in (i) we get
`:. int (xe^x)/(1+x)^2 dx`
`= e^x/(x+1) + int e^x/(x+1)^2 dx` `- int e^x/(x+1)^2 dx +C`
`=e^x/(x+1) +C`
Question: 18. `e^x ((1+sinx)/(1+cosx))`
Solution: `int e^x ((1+sinx)/(1+cosx)) dx`
`= int e^x ((1 + 2 sin.x/2 cos.x/2)/(2 cos^2 .x/2))dx`
`= int e^x (1/2 sec^2 x/2) + tan .x/2) dx`
`=1/2int e^x sec^2*x/2dx + int e^x tan * x/2 dx`
`=1/2 int e^x sec^2 * x/2 dx + I_1`
Now, `I_1 = int e^x tan*x/2dx`
Taking `tan*x/2` as first function and `e^x` as second function then
`int e^x tan*x/2 dx`
`=tan*x/2 int e^x dx -` `int[e/ex(tan*x/2) inte^x dx]dx`
`e^x tan *x/2 e^x dx - int 1/2 sec^x*x/2 e^x dx + C`
`=e^x tan*x/2 - 1/2 int e^x sec^2*x/2dx+C`
Putting value of `I_1` in (i) we get
`:. int e^x((1+sinx)/(1+cosx))dx`
`=1/2int e^x sec^2*x/2dx + e^x tan*x/2-` `1/2int e^x sec^2*x/2 dx+C`
`e^x tan *x/2+C`
Question: 19. `e^x(1/x - 1/x^2)`
Solution: `int e^x(1/x - 1/x^2)dx`
`=int e^x 1/x dx - int e^x 1/x^x dx`
`=I_1 - int e^x 1/x^2 dx` ----(i)
Now, `I_1 =e^x 1/x dx`
Taking `1/x` as first function and `e^x` as second function then
`I-1 = 1/x int e^x dx-` `int [e/dx(1/x)int e^x dx]dx`
`=1/x e^x dx - int-1/x^2 e^x dx`
`=e^x/x+int e^x 1/x^2dx +C`
Putting the value of `I_1` in eqution (i), we get
`:. int e^x (1/x-1/x^2)dx`
`=e^x/x+int e^x 1/x^2 dx` `int e^x 1/x^2 dx+C`
`=e^x/x +C`
Question: 20. `(x-3)/(x-1)^3 e^x`
Solution: `int (x-3)/(x-1)^3 e^x dx`
`=int (x-1-2)/(x-1)^3 e^x dx`
`=int [1/(x-1)^2 - 2/(x-1)^3]e^x dx`
`=int e^x 1/(1-x)^2 dx-2 int e^x 1/(x-1)^3dx`
`=I_1 - 2 int e^x 1/(x-1)^3dx` ----(i)
Now, `I_1=inte^x 1/(1-x)^2 dx`
Taking `1/(x-1)^2` as first funcion and `e^x` as second function then
`I_1 = 1/(1-x)^2 int e^x dx -` `int [d/dx(1/(x-1)^2)int e^x dx]dx`
`=e^x/(x-1)^2 - int (-2)/(-1)^3 e^x dx+C`
`=e^x/(x-1)^2-int (-2)/(x-1)^3 e^x dx+C`
Putting the value of `I_1` in (i) we get
`:. int (x-3)/(x-1)^3 e^x dx `
`= e^x/(x-1)^2 +2int e^x 1/(x-1)^3 dx -` `2int e^x 1/(x-1)^3 dx`
`=e^x /(x-1)^2 +C `
Question: 21. `e^(2x) sin x`
Solution: `int e^(2x) sin xdx`
Taking `sin x` as first function and `e^(2x)` as second function
`:. int e^(2x) sin xdx`
`=sinx int e^(2x) dx -` `int [d/dx (sinx) int e^(2x)dx]dx``=sin x e^(2x)/2-int cosx e^(2x)/2 dx`
`=sin x e^(2x)/2-1/2int e^(2x) cos x dx`
`=sin x e^(2x)/2-1/2I_1` ----(i)
Now `I_1=int e^(2x)cosx dx`
Taking `cos x` as first function and `e^(2x)` as second funtion then
`I_1=cosx int e^(2x)dx -` `int [d/dx(cosx)int e^(2x)dx]dx`
`=cos x e^(2x)/2 dx -` `int -sin x e^(2x)/2 dx +C_1`
`=(e^(2x)cosx)/2 +1/2int e^(2x) sin x dx + C_1`
Putting value of `I_1` in equation (i) we get
`int e^(2x) sin x dx `
`=(e^(2x)sinx)/2 -` `1/2[(e^(2x) cos x)/2+1/2int e^(2x) sin x dx+C_1]`
`=(e^(2x)sinx)/2 - (e^(2x)cosx)/4 -` `1/4int e^(2x) sin x dx - 1/2 C_1`
`=>int e^(2x) sin x dx + 1/4 int e^(2x) sin x dx`
`=(e^(2x) sin x)/2 - (e^(2x)cos x)/4 - 1/2 C_1`
`=>5/4 int e^(2x) sin x dx` `=(e^(2x)sinx)/2 - (e^(2x)cosx)/4 - 1/2C_1`
`=>int e^(2x) sin x dx` `= 4/5[(e^(2x)sinx)/2-(e^(2x)cosx)/4-1/2C_1]`
`=2/5 e^(2x) sin x - 1/5 e^(2x) cosx - 2/5 c_1`
`=2/5 e^(2x) sin x - 1/5 e^(2x) cos x + C`
[Where `C=-2/5C_1`]
`=e^(2x)/5 (2sinx - cos x)+C`
Question: 22. `sin^-1((2x)/(1+x^2))`
Solution: `intsin^-1((2x)/(x+x^2))dx`
Put ` x = tan t => dx = sec^2 tdt`
`:. int sin^-1((2x)/(1+x^2))dx`
`=int sin^-1((2tant)/(1+tan^2t))sec^2tdt`
`=int sin^-1 (sin2t) sec^2tdt`
`=int 2t sec^2t dt`
`=2int t* sec^2 t dt`
Taking t as first function and sec2t as second function then
`2intt*sec^2 t dt`
`=2[t*intsec^2 t dt-` `int [d/dx(t)*intsec^2tdt]dt]`
`=2[t tant - int tant dt]`
`=2t tan t - 2 log|sec t|+C`
`=2t tan t - 2 log(1+tan^2 t)^(1/2) + C`
`:.int sin^-1((2x)/(1+x^2)) dx`
`=2x tan^-1 x - 2log sqrt(1+x^2) +C`
Choose the correct answer in Exercise 23 and 24
Question: 23. `int x^2 e^(x^3) dx` equals
(A) `1/3 e^(x^3) +C`
(B) `1/3 e^(x^2) +C`
(C) `1/2 e^(x^3) +C`
(D) `1/2 e^(x^2) +C`
Answer: Option (A) `1/3 e^(x^3) +C` is the correct answer
Explanation: `int x^2 e^(x^3) dx`
Put, `x^3=t=>3x^2 dx=dt`
`=> x^2 dx = 1/3 dt`
`:. int x^2 e^(x^3)dx`
`=1/3 int e^t dt`
`=1/3 e^t+C`
`=1/3e^(x^3)+C` Answer
Question: 24. `int e^x sec x(1+tan x) dx` equals
(A) `e^x cos x+c`
(B) `e^x sec x+c`
(C) `e^x sin x+c`
(D) `e^x tan x+c`
Answer: Option (B)`e^x sec x+c`
Explanation:
`int e^x sec x(1+tan x) dx`
`=int e^x sec x dx +e^xsec x tan x dx`
`=I_1+int e^x sec x tan x dx`
Now, `I_1 = int e^x sec x dx`
Taking sec x as first function and `e^x` as second function then
`I_1=sec x int e^x dx -` `int[d/dx(secx)int e^xdt]`
`=secx e^x dx-` `int sec x tanx e^x dx`
`e^x sec x -` `int e^x sec x tanx dx +C`
Putting value of `I_1` in (i), we get
`int e^x sec x (1+tanx+c)dx`
`=e^x secx -` `e^x sec x tan x dx` `int e^x sec x tan x dx +C`
`= e^x sec x +C` Answer
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