第12类RD Sharma解-第19章不定积分–练习19.12

问题^{1∫sin4×COS 3×DX}

解决方案：

Let I = ∫ sin⁴x cos³x dx          -(i)

Let sinx = t

On differentiating with respect to x:

cosx = dt/dx

cosx dx = dt

dx = dt/cosx

Putting value of dx and sinx in equation (i):

 I = ∫ t⁴ cos^xdt/cosx
 I = ∫ t⁴ cos² x dt

 I = ∫ t⁴ (1 – sin² x) dt

 I = ∫ t⁴ (1 – t²) dt

 I = ∫ (t⁴– t²) dt

 I  = t⁵/5 – t⁷/7 + c

I  = sin⁵/5 – sin⁷/7 + c

问题2。∫sin ⁵ x dx

解决方案：

Let I = ∫ sin⁵x dx 

I = ∫sin³xsin²x dx

= ∫sin³x(1 – cos²x)dx

= ∫(sin³x – sin³xcos²x)dx

= ∫[sinxsin²x – sin³xcos²x]dx

= ∫[sinx(1 – cos²x) – sin³xcos²x]dx

= ∫(sinx – sinxcos²x – sin³xcos²x)dx

I = ∫sinx dx – ∫sinxcos²x dx – ∫sin³xcos²x dx

Putting cosx = t and -sinxdx = dt in 2nd and 3rd integral:

I = ∫sinx dx + ∫t²dt + ∫sin²xt³dt/t

= ∫sinx dx + ∫t² dt + ∫sin²xt² dt​

= ∫sinx dx + ∫t² dt + ∫(1 – cos²x)t² dt

$= -cosx+ \frac{t^3}{3} + \int (1-t^2)t^2dt\\ = -cosx + \frac{t^3}{3} + \int(t^2-t^4)dt\\ -cosx+\frac{t^3}{3}+\frac{t^3}{3}-\frac{t^5}{5}\\ -cosx+\frac{2t^3}{3}+\frac{t^5}{5}+c$
Putting value of t:
$\\ I=-cosx+\frac{2cos^3}{3}+\frac{cos^5}{5}+c$

问题3。   ^∫cos5×DX

解决方案：

Let I = ∫cos⁵x dx

I = ∫cos²xcos³x dx

= ∫(1 – sin²x)cos³x dx

= ∫(cos³x−sin²xcos³x)dx

= ∫(cos²xcosx – sin²xcos²xcosx)dx

= ∫[(1 – sin²x)cosx – sin²x(1 – sin²x)cosx]dx

= ∫(cosx – sin²xcosx – sin²xcosx + sin⁴xcosx)dx

= ∫cosx dx – 2∫sin²xcosx dx + ∫sin⁴xcosx dx 

Putting sinx = t and cosxdx = dt in 2nd and 3rd integral we get:

I = ∫cos dx – 2∫t²dt + ∫t⁴dt

= sinx – 2t³/3 + t⁵/5 + c

Putting value of t:

I = = sinx – 2sin³x/3 + cos⁵x/5 + c 

问题4。   ^∫sin5 xcosx DX

解决方案：

Let I = ∫sin⁵xcosx dx         −(i)

Let sinx = t:

On differentiating with respect to x:

-cosx = dt/dx

cosx dx = -dt

Putting cosxdx = -dt and sinx = t in eq (i):

I = ∫t⁵dt

= t⁶​/6 + c

= sin⁶​x/6 + c

问题5   ^∫sin3 xcos ^6×DX

解决方案：

Let I = ∫sin³xcos⁶x dx           −(i)

Let cosx = t

On differenciating both sides w.r.t′x′:

-sinx = dt/dx

sinxdx = -dt​

Putting cosx = t and sinxdx = -dt in eq (i):

I = -∫sin²x t⁶dt

= -∫(1 – cos²x)t⁶dt

= -∫(1 – t²)t⁶dt

= -∫(t⁶ – t⁸)dt

= -(t⁷/7​ – t⁹/9​) + c

Putting value of t:

I = -(cos⁷x/7​ – cos⁹x/9​) + c

问题6。   ^∫cos7×DX

解决方案：

Let I = ∫cos⁷x dx

= ∫cos⁶xcosx dx

= ∫(cos²x)³cosx dx

= ∫(1 – sin²x)³cosx dx

= ∫(1 – sin⁶x – 3sin²x + 3sin⁴x)cosx dx

= ∫(cosx – sin⁶xcosx – 3sin²xcosx + 3sin⁴xcosx)dx         −(i)

Putting sinx = t and cosx dx = t in 2nd,3rd and 4th integral in (i):

I = ∫cosx dx – ∫t⁶dt – 3∫t²dt + 3∫t⁴dt

= sinx – t⁷/7 ​- 3t³/3 ​ +3t⁵/5​ + c

Putting value of t:

= sinx – sin⁷x/7 ​- 3sin³x/3 ​ +3sin⁵x/5​ + c

问题7。   ^∫xcos3×2的SiNx ² DX

解决方案：

Let I = ∫xcos³x²sinx²dx          −(i)

Let cosx² = t

On differentiating both sides:

 -2xsinx² = dt/dx

​ xsinx² dx = –dt/2

Putting values in (i):

$I=\int t^3\frac{-dt}{2}\\$

= -t⁴​/8 + c

Putting value of t:

$I=-\frac{1}{8}cos^4x^2+c$

问题8。   ^∫sin7×DX

解决方案：

Let I = ∫sin⁷x dx

I = ∫sin⁶x sinx dx

= ∫(sin²x)³sinx dx

= ∫(1 – cos²x)³sinx dx

= ∫(1 – cos⁶x – 3cos²x + 3cos⁴x)sinx dx

I = ∫sinx dx – ∫cos⁶xsinx dx + 3∫cos⁴xsinx dx – 3∫cos²xsinx dx

Putting cosx = t and sinx dx = -dt in 2nd,3rd and 4th integral:

I = ∫sinx dx – ∫t⁶(-dt) + 3∫t⁴(-dt) – 3∫t²(-dt)

$=\ -cosx+\frac{t^7}{7}-\frac{3}{5}t^5+\frac{3}{3}t^3+c\\ =\ -cosx+\frac{cos^x}{7}-\frac{3}{5}cos^5x+cos^3x+c\\ \implies -cosx+cos^3x-\frac{3}{5}cos^5x+\frac{1}{7}cos^7x+c$

问题9。   ^∫sin3 xcos ^5×DX

解决方案：

Let I = ∫sin³xcos⁵x dx         −(i)

Let cosx = t

On differentiating both sides: -sinx = dt/dx

sinx dx = -dt​

Putting values in (i):

I = ∫sin²xt⁵(-dt)

= ⁻∫(1 – cos²x)t⁵ dt

= ⁻∫(1 – t²)t⁵ dt

= ∫(t⁷ – t⁵) dt

= t⁸/8 – t⁶/6​ + c

Putting value of t:

$\implies \frac{1}{8}cos^8x-\frac{1}{6}cos^6x+c$

问题10。 $\int \frac{1}{sin^4xcos^2x}dx$

解决方案：

Let I = $\int \frac{1}{sin^4xcos^2x}dx\quad -(i)$

Dividing and multiplying the equation by cos⁶x:

$I=\int \frac{\frac{1}{cos^6x}}{\frac{sin^4xcos^2x}{cos^6x}}\\ =\ \int \frac{sec^6x}{tan^4x}dx\\ =\ \int \frac{sec^4xsec^2x}{tan^4x}dx\\ =\ \int \frac{(sec^2x)^2sec^2x}{sec^2x}dx\\ =\ \int \frac{(1+tan^2x)^2sec^2x}{tan^4x}dx\\ I=\int \frac{(1+tan^4x+2tan^2x)sec^2x}{tan^4x}dx\quad -(ii)\\$

Let tanx = t, then:

sec²x = dt/dx

sec²x dx = dt

Putting values in eq (ii):​

$\\ I=\int \frac{1+t^4+2t^2}{t^4}dt\\ =\ \int (\frac{1}{t^4}+1+\frac{2}{t^2})dt\\ =\ -\frac{1}{3t^3}+t-\frac{2}{t}+c\\ =\ -\frac{1}{3tan^3x}+tanx-\frac{2}{tanx}+c\\ \implies -\frac{1}{3}cot^3x-2cotx+tanx+c$

问题11  $\int \frac{1}{sin^3xcos^5x}dx$

解决方案：

$Let\ I=\int \frac{1}{sin^3xcos^5x}dx$
Dividing and multiplying by cos⁸x:
$\\ =\ \int \frac{\frac{1}{cos^8x}}{\frac{sin^3xcos^5x}{cos^8x}}dx\\ =\ \int \frac{sec^8x}{tan^3x}dx\\ =\ \int \frac{(sec^2x)^3}{tan^3x}sec^2xdx\\ =\ \int \frac{(1+tan^2x)^3}{tan^3x}sec^2xdx\\ =\ \int \frac{(1+tan^6x+3tan^2x+3tan^4x)sec^2x}{tan^3x}dx$
Let tanx=t,then:
$\\ sec^2x=\frac{dt}{dx}\\ \implies sec^2xdx=dt$
Putting values in ii:
$\\ I=\int \frac{1+t^6+3t^4+3t^2}{t^3}dt\\ =\ \int (\frac{1}{t^3}+t^3+3t+\frac{3}{t})dt\\ =\ \frac{1}{2t^2}+\frac{t^4}{4}+\frac{3t^2}{2}+3logt+c\\ \implies I=\frac{-1}{2tan^2x}+3log|tanx|+\frac{3}{2}tan^2x+\frac{1}{4}tan^4x+c$

问题12。  $\int \frac{1}{sin^3xcosx}dx$

解决方案：

$Let\ I=\int \frac{1}{sin^3xcosx}dx$
Dividing and multiplying by cos⁴x:

$\\ I=\int \frac{\frac{1}{cos^4x}}{\frac{sin^3xcosx}{cos^4x}}dx\\ I=\int \frac{sec^4x}{tan^3x}dx\\ I=\int \frac{sec^2xsec^2x}{tan^3x}dx\\ =\ \int \frac{1+tan^2x}{tan^3x}dx \quad -i$
Let tanx=t,then:
sec²xdx = dt
Putting values in i:
$\\ I=\int \frac{1+t^2}{t^3}dt\\ I=\int (\frac{1}{t^3}+\frac{1}{t})dt\\ =\ -\frac{1}{2t^2}+log|t|+c$
Putting value of t:
$\\ \implies -\frac{1}{2tan^2x}+log|tanx|+c$

问题13。  $\int \frac{1}{sinxcos^3x}dx$

解决方案：

$Let\ I=\int \frac{1}{sinxcos^3x}dx\\ \frac{1}{sinxcos^3x}= \frac{sin^2x+cos^2x}{sinxcos^3x}\\ =\ \frac{sinx}{cos^2x}+\frac{1}{sinxcosx}\\ =\ tanxsec^2x+\frac{\frac{1}{cos^2x}}{\frac{sinxcosx}{cos^2x}}\\ =\ tanxsec^2x+\frac{sec^2x}{tanx} \implies \int \frac{1}{sinxcos^3x}dx= \int tanxsec^2xdx +\int \frac{sec^2x}{tanx}dx$
Let tanx=t⟹sec²x dx = dt:
$I= \int tdt+\int \frac{1}{t}dt\\ =\ \frac{t^2}{2}+log|t|+c$
Putting value of t:
$\implies I=\frac{1}{2}tan^2x+log|tanx|+c$