第12类RD Sharma解–第19章不定积分–练习19.13 |套装2

问题11.评估∫sin2x ^{/√cos4×-SIN 2×+} 2 DX

解决方案：

Let us assume I =∫ sin2x/ √cos⁴x-sin²x+2 dx

=∫ sin2x/ √cos⁴x-(1-cos²x)+2 dx (i)

Put cos²x = t

-2sinxcosx dx = dt

sin2x dx = -dt

Put the above value in eq. (i)

= -∫ dt/ √t²-(1-t)+2

= -∫ dt/ √t²-1+t+2

= -∫ dt/ √t²+t+1

= -∫ dt/ √t²+t+(1/4)+(3/4)

= -∫ dt/ √(t+1/2)²+ 3/4 (ii)

Put t+1/2 =u

dt = du

Put the above value in eq. (ii)

= -∫ du/ √u²+ 3/4

= -∫ du/ √u²+3/4

Integrate the above eq. then, we get

= -log|u +√u²+3/4| + c [since ∫ 1/√x²+a² dx =log|x +√x²+a²| + c]

= -log|t+1/2 +√(t+1/2)²+3/4| + c

= -log|t+1/2 +√(t²+t+1)| + c

= -log|(cos²x+1/2) +√(cos⁴x+cos²x+1| + c

Hence, I = -log|(cos²x+1/2) +√(cos⁴x+cos²x+1| + c

问题12评估∫cosx /√4-SIN ^2×DX

解决方案：

Let us assume I =∫ cosx/ √4-sin²x dx (i)

Put sinx = t

cosx dx = dt

Put the above value in eq. (i)

= ∫ dt/ √(2)²-(t)²

Integrate the above eq. then, we get

= sin^-1(t/2) + c [ since ∫1/ √a² – x² dx = sin^-1(x/a) + c]

= sin^-1(sinx/2) + c

Hence, I = sin^-1(sinx/2) + c

问题13.评估∫1 ^{/ x 2/ 3√x2/3} -4 dx

解决方案：

Let us assume I =∫ 1/ x^2/3√x^2/3-4 dx (i)

Put x^1/3 = t

(1/3) x^1/3-1 dx = dt

(1/3) x^-2/3 dx = dt

dx/ x^2/3 = 3dt

Put the above value in eq. (i)

= 3 ∫ dt/ √t²-(2)²

Integrate the above eq. then, we get

= 3 log|t +√t²-(2)²| + c [since ∫ 1/√x²-a² dx =log|x +√x²-a²| + c]

= 3 log|x^1/3 +√(x^1/3)²-(2)²| + c

Hence, I = 3 log|x^1/3 +√x^2/3-4| + c

问题14.评估∫1 /√(1-x ² )[9+(sin ^-1 x) ² dx

解决方案：

Let us assume I =∫ 1/ √(1-x²)[9+(sin^-1x)² dx (i)

Put sin^-1x = t

dx/√1-x² = dt

Put the above value in eq. (i)

= ∫ dt/ √(3)²+t

Integrate the above eq. then, we get

= log|t +√(3)²+t²| + c [since ∫ 1/√a²+x² dx =log|x +√a²+x²| + c]

= log|sin^-1x +√(3)²+(sin^-1x)²| + c

Hence, I = log|sin^-1x +√9+(sin^-1x)²| + c

问题15.评估∫cosx ^{/√sin2×-2sinx-3} DX

解决方案：

Let us assume I =∫ cosx/ √sin²x-2sinx-3 dx (i)

Put sinx = t

cosx dx = dt

Put the above value in eq. (i)

= ∫ dt/ √t²-2t-3

= ∫ dt/ √t²-2t+(1)²-(1)²-3

= ∫ dt/ √(t-1)²-(2)² (ii)

Put t-1 =u

dt = du

Put the above value in eq. (ii)

= ∫ du/ √u²-(2)²

Integrate the above eq. then, we get

= log|u +√u²-(2)²| + c [since ∫ 1/√x²-a² dx =log|x +√x²-a²| + c]

= log|t-1 +√(t-1)²-4| + c

= log|t-1 +√t²-2t+1-4| + c

= log|t-1 +√t²-2t-3| + c

= log|sinx-1 +√sin²x-2sinx-3| + c

Hence, I = log|sinx-1 +√sin²x-2sinx-3| + c

问题16.评估∫√cosecx-1dx

解决方案：

Let us assume I =∫ √cosecx-1 dx

= ∫ √1/sinx -1 dx

=∫ √1-sinx /sinx dx

=∫ √(1-sinx)+(1+sinx) /(1+sinx)sinx dx

=∫ √(1+sinx-sinx-sin²x) /sin²x+sinx dx

=∫ √cos²x /sin²x+sinx dx

=∫ cosx /√sin²x+sinx dx (i)

Let sinx = t

cosx dx = dt

Put the above value in eq. (i)

= ∫ dt/√t²+t

= ∫ dt/√t²+2t(1/2)+(1/2)²-(1/2)²

= ∫ dt/√(t+1/2)²-(1/2)² (ii)

Let t+1/2 = u

dt = du

Put the above value in eq. (ii)

= ∫ dt/√(u)²-(1/2)²

Integrate the above eq. then, we get

= log|u +√u²-(1/2)²| + c [since ∫ 1/√x²-a² dx =log|x +√x²-a²| + c]

= log|t+1/2 +√(t+1/2)²-(1/2)²| + c

= log|t+1/2 +√t²+t| + c

Hence, I = log|sinx+1/2 +√sin²x+sinx| + c

问题17.评估∫sinx-cosx /√sin2xdx

解决方案：

Let us assume I =∫ sinx-cosx/ √sin2x dx

∫ sinx-cosx/ √sin2x dx = ∫ sinx-cosx/ √(sinx+cosx)² -1 dx

= ∫ sinx-cosx/ √(sinx+cosx)² -1 dx (i)

Let sinx+cosx = t

cosx-sinx dx = dt

Put the above value in eq. (i)

= -∫ dt/ √t²-(1)²

Integrate the above eq. then, we get

= – log|t +√t²-(1)²| + c [since ∫ 1/√x²-a² dx =log|x +√x²-a²| + c]

= – log|sinx+cosx +√sin2x| + c

Hence, I = – log|sinx+cosx +√sin2x| + c